Permutations and Combinations Lecture

hello everyone welcome back to your favorite Channel this is Gautam me mathematics Master teacher and I'm back with another exciting session so guys which is permutations and combinations let us start this session today okay so first permutations and combinations [Music] permutations and combinations and combinations and combinations foreign different type of problems a good number of let's say Okay 20 or minimum 20 30 problems yes I session look a problem what is the approach I'm following for the questions permutations and combination permutations and combinations permutations and combinations foreign we are going to discuss because six topics fundamental principle of counting very special problems like the arrangement problems and distribution of objects so six topics with the usual the questions a competitive examination E6 topics questions so let us understand in a detailed way permutation and combination so first guys which is fundamental principle of counting low fundamental principle of multiplication so for example let us say let us say foreign [Music] let us suppose foreign [Music] foreign okay this is like kind of real life scenario right from your general College foreign [Music] foreign [Music] of course foreign foreign foreign foreign I can give you another example let us saying for example let us say for example is okay let us say uh Hyderabad foreign [Music] foreign [Music] foreign foreign and the answer is again going to become 6 so this is the fundamental principle of counting low multiplication concept and a b multiplied same thing is given for you okay now let us quickly do a few questions practice key okay funny questions number of parts from A to B are four from B to C are three then find the number of paths three three a to c why are B how many number of B's you're gonna get in this case 4 into 3 and the answer is simple foreign [Music] foreign foreign foreign [Music] foreign multiply 7 into 6 into 5 is going to give you the answer so 6 5 30 into 7 210 there you go we answered so I hope this question is clear for you guys okay in fact questions which is fundamental principle of addiction multiplication let us suppose let us suppose um again of three options foreign foreign foreign foreign room foreign can you tell me any number of ways low foreign foreign foreign foreign so I hope it's clear for you guys okay foreign foreign okay so I'm just telling you formula major like theorem is generally so let us see suppose you know which I do find the number of four letter words with or without meaning which can be formed out of the letters of the word make with the repetition of the letters is not allowed image which can be formed out of the letters of the word make foreign [Music] let us suppose first place first place [Music] foreign foreign of course one day each and every place name a four letter word law each and every letter any number of ways of course you have to multiply this so four into three into two four threes are 12 12 those are twenty four so total number of ways [Music] Common Sense okay so now let me do one foreign [Music] such that the vowels are in extreme positions extreme positions first and last position you know the first position of e particular position though and E particular positions simple first you okay so basically a particular position any waste of LGH so that is two weeks okay and you get a repetition not allowed to get a question you guys try to understand this foreign foreign [Music] so I hope this is clear for you guys don't get confused if you do four into three into two into one here it has four factorial of course right the factorials product of first and natural numbers slope n is always going to belong to the positive integers including 0. okay it is going to belong to positive integers until natural numbers including 0 and remember zero factorial value which is only okay so I hope this is clear for you later the month okay how many vowels sorry how many words can be formed rearranging the letters of this Delhi such type of woman's occupy even places so basically Delhi low okay you want to do um how many words we can get such that vowels occupy even places even places so this is an even place this is so that is and you have to fill this tree this is very simple three ways we need two ways so what is the answer sorry 2 into so basically multiply since you need to find out the entire word so 2 into 3 into 2 into 1 the last question three factorial so two into three factorial is going to be that sure so option A is going to be the right answer for this question so I hope it is so now let me go to another question for the question how many four letter words four letter words containing G can be formed using the letters of daughter and in order conditions repetition not allowed reputation a lot so a daughter only let us use four letter word forms only foreign [Music] G has four options foreign foreign foreign foreign [Music] foreign guys [Music] foreign foreign according to the question [Music] I'll remove all the words in which G is not contained the total number of words flow any words of words how many number of words do we get in which G is not there very simple only four letter words the total number of words in which G is not present and ultimately 7 raised to four seven raised to 4 8 raised to four which is number of words in which G is not contained and things are uncondi whatever the remaining result you're gonna get words so I hope this question is clear for you guys so reputation allowed first reputation questions okay so I hope this is clear for everyone let's do a question how many three letter words can be made using the letters of the word daughter such that if the word contains G then it must contain H also foreign G is selected so Jesus is [Music] two ways hello you can say it can be filled in two ways so G the two ways actually it's not two ways G converter okay so let us suppose foreign foreign [Music] six [Music] six twelve eighteen twenty four Thirty and thirty six it was clear for you guys six six six six different number of busy one 6 plus 6 12 and even in different cases evenly okay you should not be multiplying it is maybe so six plus six twelve is six eighteen twenty four Thirty thirty six or two minutes hours G not a selected foreign [Music] foreign [Music] 246 let us go to some problems or numbers [Music] and combination chapter law numbers where the questions of style like certain digits will start will say that a certain digit store D five digit number or four digit form number forms a such that it is divisible by 3 divisible by 2 divisible by five different different ways foreign [Music] how many four digit not possible repetition of digits can be made with no digit less than five basically by four digit number such that digit liquids and digits can be repeated okay but lower digit should be less than no digit should be less than five so what does that means anything zero one two three and four e digits use shares Colony you need to basically form a four digit number forms place so this can be filled once placed foreign [Music] foreign foreign 125 into 4 that is 500 so I hope this is clear for everyone out there okay with every digit with every digit less than 5. okay but you get the logic right I no digits eight and nine of this one six seven eight nine okay so I hope this is clear for everyone in fact so digits more than five and a particularly six is seven eight and nine okay so I hope this is clear for everyone out there okay so now guys sorry how many four digit numbers can be formed using one two three using one two three how many four digit numbers can be formed use JC four digit numbers definitely the repetition is allowed okay so we can say this can be filled in three bits can be filled in three three ways and three days so the answer should be 3 raised to four simple okay so nothing complicated in this particular question next question given the digits one two three four foreign how many three digit numbers less than 500 can be formed if each digit is used at most once second how many of those are divisible by five okay the first place observed any question the first question is one two three four five nine number three digit number forms how many three digit numbers less than less than 500 can be formed if each digit is used at most once okay okay guys so you can take only this number so this position can be filled in four weeks foreign foreign the right answer for this question is going to become 80. so I hope it's legal for you first place one two three foreign how many of these are divisible by five how many of these are there one two three and four and a four ways using those digits only foreign foreign foreign going to become so is this clear for everyone out there simple previous question learning first position is final positions which is going to be 16 okay so I hope this is clear for you in fact okay how many three digit numbers less than 600 and divisible by 5 can be formed using 0 1 2 3 4 5 9.3 because less than 600 so I hope you guys can do this strategy question okay so now let me go to another question it already changes let's go to this question guys how many four digit numbers can be formed using one two three four which are divisible by four which are divisible by four divided by four antioxide what is the divisibility test for four guys divisibility test for 480 last two digits should be divisible by four column so until it is so you've been fixed value 24 2 square 24 32 this question 32 this version so one two three twelve uh 24 32 until so basically places you know three ways of three ways 12 24 32 foreign guys [Music] uh here the repetition is allowed how many four digit numbers can be formed using one two three four which are divisible by which are divisible by four so e k slow we can take reputation is allowed because I mentioned the other question right reputation I lovedi then we can say that this can be filled in four ways right this can also be filled in four ways okay so 4 4 16 and last number 241 21 23 24 uh 41 okay 31 32 33 34 41 42 43 44 perfect so what is the difference maybe I might be making some mistake in this case okay Mission guys edit 12 . sorry sorry I think uh I get it application sorry blue anime so let's suppose he named 12th of illness is low if you fill the last two digits by 12. let us suppose which one with 6 12 30 to 24th this is how you get it adjourned up sorry 12 24 32 and half 44 foreign [Music] right so if you do it in that way then clearly see that means four ways to repeat GH 0 and the remaining two places can be taken in four into four which is going to be uh definitely uh 16 ways the 16 into 4 64 answer key right so options 64 in the 64 option level 125 yeah sorry sorry I think there might be something uh wrong options so answer is going to become 64. sixteen fours are it's going to become 64 not 125 the right answer for this question is 64. okay so simply right I hope this is clear for you guys so for divisibility test tells in the last two digits so now let's go to another question in fact a question let me tell you one important point a point question is actually uh let us say you have to form a three digit number under 2 which is either divisible by 2 or 300 either divisible by two or three same question divisible by two or three energy so a lot go with this formula this and this is n of B minus n of a intersection sorry so equation you want to look sorry how many four digit numbers can be formed from which either begin with 3 or end with 8. how many four digit numbers so question number is how many four digit numbers can be formed which either begin with a or end with b a a begin with the 3 or end with eight I think let us suppose I'll consider this a as the number of ways in which this particular number okay is going to start with 3m so a or J C the total number of ways foreign how many how many four digit numbers can be formed which either begin with three or end with eight okay so in your sorrow we make it up three places slow or three or less three places [Music] the first and last position to 0 3 and 8th of agent a intersection b and a and t first three last eight under three guys got it is [Music] right so I hope this is here for you this is how you're gonna get the answer okay so that is so I needed this quantity so this can be found in 10 ways 10 waves and 10 ways so enough a 10 Cube also okay okay so this can be filled in 10 ways in 10 digits firstly 10 ways to fill J alone Square nine into ten Square and what is n of a intersection B another intersections so run the place in the middle foreign foreign what did we take a as a and a in this start with three but a complementary no it doesn't start with three so B and 10 is going to start ends with eight so we complemented not ending with it so a complement intersection B complement which is n of a union B whole complement the D Morgan is nothing but total the total of chasing you already know this n of a union B value substitute Json B you are going to get done so so in the equation under e so this is how you have to do it guys simple question either divisible by this or this is foreign s which are you know very helpful for us in counting year two results the number of ways of arranging are distinct objects at our places the number of ways of arranging are distinct objects that are clicks four places four objects online for four ways foreign foreign okay in fact we can also discuss the concept of permutation guys permutation is angular Arrangements foreign [Music] foreign [Music] foreign so I hope this is clear for all the students out there okay fine let us get back to this one if you don't talk ceremony number of ways of selecting you grow Arrangement typing Arrangement key selection key difference only you know order important let us say Okay um order report foreign 40 members low foreign so basically selection the order is not going to be important so remember that so okay let us suppose n objects and distinct objects and distinct objects are number of objects the formula is n factorial by R factorial into n minus our whole factorial in the formula right in fact foreign I think I mean the case of arrangement the order is going to matter of the way order matter of something first ever second level third episode [Music] order then long in the case of the order that's not right order which is an order does not matter in the case of which one in the case of selection so e Arrangements we are calling it as permutation the arrangements permutation and the selection on the combination permutations and combinations permutations and combinations basically Arrangements so I hope this is clear for everyone out there okay foreign foreign thank you foreign foreign foreign [Music] foreign [Music] you can start with selecting four men first case Okay selecting four men four men of course only one woman five members committee is one woman so how many ways can we do this so basically six men low four months anyway and five women select German H1 then three women three women so three women selectors so three women and two men okay so three women from five womens five c three and two men Allah from six men six feet so these are the two different so what is going to be the answer is if you add this whatever the result you're gonna get that is going to be the answer so got it simple question so four minutes containing four men and three women 64 into five three five guys I think uh the question below first question low okay I'll tell you the difference by a committee of five members five members foreign so I hope this is Cleo for every one of the requesting Theory this is going to be the answer okay this is what you want to get okay and coming to the next question how many committees contain at most two women at most two men sorry women so let us suppose you get a question within five members okay restriction Zero menu and five moment of Star Trek that is one scenario or else one man and four women of Sanitation or else Two Men and three women who started committee right and we can't go more than two because at most two men in order so what is how many ways can we select this on how many ways can selection be done in this case so 0. has to be selected which is 5 C 4 and this one two minutes so 62 into three women select Journey five women windshield it's going to be five c three which is to end up getting the clear I hope this is clear for you simple question is you have to get the answer as 181 okay so fine now let us go to one thing let us do one thing guys let us go to the next question how many committees contain at least one woman at least one woman one woman and four men the five members only so one woman four men presented two women Three Men and three women two men isn't a four women one man listen to five women so this in this many ways I can select a committee so the first one over men's energy so 64 into 5c one lady next three men sixty three into five C two will be foreign so I hope this question is clear for all the students out there depending upon the criteria given for you in the question so I hope this is clear for you so now let me go to another question a group of students consisting of four boys and five girls find the number of ways of selecting a team of at least three boys and four girls can you guys do it find the number of ways of selecting a team of at least three boys and four girls so at least a group of friends consists of four boys and five girls okay find the number of ways of selecting a team of at least three boys and four words so at least three boys boys and girls slow minimum three so how many ways in how many ways can you select three boys and four girls together so it's going to be four C three spicy forever next you can also go with four and five so minimum at least one day you can also go with four boys as well as five girls or do you know four boys and four girls at least three boys and four girls four boys four girls could have corrected in that case you get four C four into five C four or you can set three boys and five girls yes that is also Pro characters at least three at least four so this is like four C three into five C five five simple and finally you can go with four points and five months four points and finals and that is four C four into five C five just add them away whatever result you are gonna get that is going to be the answer okay so I hope this is clear for everyone only simple question is you can easily understand how to do the selection in this particular case you can see here the same thing one two three and the fourth case okay so now let me go to the next one Jee main 2020 question questions there are three sections in a question paper okay so let us see that section 1 Section 2 section three okay and each section contains five questions okay section one of five questions is section two or five questions so section three a lot of five questions are fine a candidate has to answer a total of five questions choosing at least one question from each section okay in the number of ways in which a candidate can choose the questions at least one uh questions foreign try to understand this I can do it in this way see this is about the number of ways you can pick up here so I can say that okay okay um what we will do is uh let's decide let's decide think about it is so that is going to be in how many number of ways three ways are the three ways so now you can say five c one five c one into five c three this is one way plus so that is why I'm telling you again this is in three weeks into monthly five c one into Phi C2 into Phi CO2 so if you do this you are going to get done so simplify J and D children that is going to give you the answer so you can see here clearly same thing 3 into 5 c 1 5 c 1 5 C 3 3 into 5 C one five six two five C two simplify this are unconding the answer is 2 to 5. so I hope this question is clear for everyone okay so first changes foreign blue the kind of cricket team ongoing a cricket team low okay players so how many ways we can select it 14 C11 so I got it everywhere so a particular player has never chosen Nikita players so 14c 11 is going to be the answer for this okay next question or not sorry three particular players are always chosen 15 players foreign foreign and eight players knee is and that is the number of ways in this particular case so got it everyone so simple I think I think is also clear for eign there are two ways of doing questions total number of ways in which you can select the team and I'll select jh2 simple you have 15 to 3 11 select General 15 C 11 select JS number one okay 15 C11 in this case also what happens is that there are number of ways the num there is number of ways in which two particular players these two players should never come together I'll remove all the cases of all the Selections in which I have selected both the players okay in how many ways can we do that guys simple the first angels are entered two players are going to come together 15 C 11 noon shall remove all the ways all the Selections in which these two players have come together 15 C 11 minus our under players okay the two particular players playing together minus two particular players together get the number of ways in which they are not together right you get the number of ways in which they are not two together so total 15 is 11 two players come together and there are two gates yes is going to be the answer so I hope this is clear for everyone okay same thing but both you end up getting the same answer guys okay so I hope it's clear for you for you so now let me do this question guys in how many ways we can select D numbers out of one two so 100 such that they form a a b they form and they are basically going to be in arithmetic combination so basically let us suppose X1 X2 and X3 these condom numbers okay such that they are they have to be in AP Allah so always just select jh123 234-345-456-678-13579 okay okay so since they are in EP what we can write is 2 x 2 is going to become X1 plus x 3 right when three numbers are in AP this is the criteria or I can say X2 is going to become x 1 plus x 3 divided by this what is clear is X1 plus X3 should be divisible by 2 yes or no X1 plus X3 is it the number because this should definitely be divisible X2 is going to be a natural number of product possible X1 plus X3 should be divisible by so how can we choose X1 and X3 so sum of two numbers divisible by 2 1 day the number should be even what is the divisibility divisibility test of two the number should be even so we know that this X One Plus X3 so we know that x 1 plus x 3 is a even number it's going to be an even number either sum of two numbers even I put out on the there are two cases for that either x one and X2 are both even you don't do even numbers we address the even number of something x 1 and X2 both are odd numbers I'm going to odd numbers you add Json odd numbers odd numbers zero you get the result as even so now you calculated how many ways you can select even numbers two even numbers out of these numbers and how many ways you can select odd numbers out of the given numbers so let us suppose so for this one so for this one for this one so any even numbers only one to hundred modulo 50. so 50 even though you don't want to select generally so how many ways okay similarly it's for an extra run to odd any odd numbers one time only 50 so 50 kilometers so 50 c two foreign [Music] so I hope this is clear for all the students out there so 2 into 50 c 2 is going to be the answer okay so munchie question will be conceptually good question so I hope you guys understood this one is [Music] here so this is can be done permission guys so this can be so let us say in the first case second case so case number one 50 c two ways case number two fifty C2 base loss this is a total interest on the graph it's going to become 2 into 50 50 c two so I hope this is clear okay so now let me go to another question in fact only method discussions okay that is okay so now let me go to this concept guys the total number of selections out of n distinct objects is 2 . and distinct objects one two three four so on and so one two three four so on and distinct objects okay so total number of selections until total number of any ways to select JH objects out of and desktop sorry out of n distinct object flow out of and distinct objects law any way slow objectives foreign foreign foreign foreign you know foreign is going to become so I hope this is clear for everyone okay so guys let's go to the questions there are three distinct physics books that four distinct mathematics Books Okay in how many ways books can be selected and the ABC questions in Shadow question number is foreign foreign services the total number of ways just add it up the 7c 0 plus 7 say one plus seven C two plus seven C seven which is nothing but sure the answer clear oh the question number eight let's go to question number B at least one book is selected at least one book is selected physics foreign [Music] foreign so I hope it's clear for you guys okay next question is selection can be made such that at least one math book is selected at least put a specifically um how many number of ways I can select the books simple okay at least one books so foreign foreign special case of selection is foreign foreign foreign in how many ways can I pick up those three chocolates out of 10 identical chocolates one way one way okay okay so from this concept which you have to understand is La n identical objects in identical objects n identical objects and identical objects okay the number of ways in which you can pick up let us say one object two objects anyways three objects anyways so on um the number of ways of selecting two A's out of Four A's is what [Music] selecting let us say zero or more objects selecting is low n identical and identical objects in identical and identical objects in identical objects in G if you want to select okay if you want to select zero or more objects select 0 or more objects here or more objects so zero or more objects this is nothing but n plus one the number of a is n plus one identical object slow object selected selection n number of times it is n foreign a foreign foreign three letters selection I get three different letters three different letters three letters different guys what are the different cases three letters and ten let us say more distinct number three is foreign next two two distinctive sorry two I like one distinction one distinct number okay so how can we how many number of faces in the world and I could have two allocated two letters foreign [Music] foreign [Music] this is going to become and finally three alike three are like anyways so this can be selected in two weeks so the total number of ways of selecting three letters six plus one three three plus two plus six plus one three guys I'm very sorry six plus one seven seven plus two nine so therefore nine ways is going to be the total of every particular case foreign foreign let's go to the next step concept and that is this question and the equation law in the application in how many ways we can select four letters out of assassination assassinations okay okay okay foreign similar to the previous question four letters four letters four distinct this question first case for a distinct any distinct letters is also four so anyways to the H is [Music] educational 64 and that is this one for distinct so four distance do I like and two distinct to alike and two distinct so two alike and that is so run the same letters you can take either two A's or two s's or two eyes or two ends four options foreign three alike wonderful three alike and one District level rights one three I like and one distinct and Rascals how can we write this down there is let us suppose a select j7 and finally 4 alike four same letters sorry only one way and the content only essay foreign you get done so if you add up all these cases you get the answer so arthma in the clear Gap I hope it's clear for you then so first thing is 64 right which is four C one into five C two four C one into five sorry four C two and I can two District Four C one one two three four five foreign and two I like to alike and type 2 una four letters will be like a a s s there is one two three four foreign for everyone out there so another two alike and two alike and three are like at one desk and already choose something it's two c one into five c one save money two c one into five c one and finally all I like is only one so if you add it up you get 72 as an answer okay so this is the question munchie questions so I hope you get it okay fine so now what we will do is let's see some geometrical counting problems counting problems selections and a combinations the first question for you guys there are end points on a circle circle final end points online first question entry joining this points how many lines will be made how many lines will be made okay so one two three four five six you have in there foreign since no point is I can write down since no point is scoring here no two points are calling here since no two points are convenient okay since no two points are convenient are cholinium what we can say in this particular case the answer the result is going to be in how many ways you can select the two points out of the given number of points that will give you the number of straight lines so this is the formula and C to another formula number of straight lines and C2 and every formula number of straight lengths when the points are given points straight lines that is okay next question in journal Theory joining this point how many triangles will be made how many ways you can select three points right since three points let's say suppose three points so since next one for you guys next one indicator joining those points of polygon is made find the number of its diagonals either number of diagonals concordance number of diagonals number of diagonals of a polygon of a polygon of enzymes of a polygon of n sites what is the formula nc2 minus l nc2 minus n or n into n minus 3 divided by 2 the formula number of diagonals concordance for n sided polygon okay so I hope this is clear for everyone out there okay so NC n into n minus 3 by 2 is going to be the formula where n is going to be the number of sides I think e particular question k n is number of points uh you do one thing guys let us say we simple the ionic equation is MC2 is going to become n e value of n substitute okay so I hope this is clear for all the students out there guys okay so this is how you have to do it uh right so save it Shadow for a diagonal which is the line we require two points but those selection will include sides of polygon also as they are all formed by selecting two points so number of diagonals of a n-sided polygon is nc2 minus n okay that the number of diagonals is n-sided polygon and nc2 minus N I hope it's clear for you okay so this is the formula covalented question ignore the same equation though equation question confusing the N points on the particular one suppose three sides questions I just cancel this question out foreign [Music] sides is going to be nc2 minus n right okay let's go to this question consider seven points four are collinear joining these points how many lines will be made first question triangles will be made you put a equation device joining this point how many lines will be made so users foreign foreign what is that formula is out of n points out of n points out of n points if M points are collinearium out of n points M points are collinear okay then then the number of straight lines the number of straight lines of a formula into simple gum and C2 minus mc2 plus 1 is a formula using that formula you can get the answer so I hope this is clear for you okay so next question is the next question is how many triangles have been made out of n points out of n points out of n points if M points are collinear okay M points are collinear then the number of then the number of which one the number of triangles the number of triangles simply nc3 minus nc3 nc3 points it will not form a triangle nc3 minus McRae a particular question answer 7c3 minus 4 CPM so I hope this is clear for everyone out there okay so now let me go to the next one text for example the question you want to talk sorry in a plane there are eight lights no two of which are parallel no three are concurrent okay how many triangles can be formed with their point of intersection as vertices foreign first number of point of intersections November of point of intersection at the point of intersections in allows in the two lines under two number of point of intersection is directly proportional and directly proportional number of intersects sorry the number of point of intersection is nothing but the number of ways of selecting two lines out of eight lines together we wrote eight lines since the lines are not parallel and no three lines are concurrent property so we can see that number of point of intersections is simply lc2 okay so what is the value so calculate the pseudometer so eight factorials by 2 factorial and 6 Factor here which is eight into seven y t so that went away 28 guys 28 number of point of intersection I think normally how many triangles can be formed under 28 points are number of triangles number of triangles number of triangles the directing are 28 C3 RS are 28 points will see three points can conserve such that you know three points three points anyways that is the number of triangles not only 28 C3 in the country 28 C3 low are you sure all the 28 points will be non-collinear foreign so any point of intersection for child three point of intersection three point of intersection which you will also get from the Formula 3 c 2 which is still I think it could absorb oxide what's the point of intersections any three are three can I say all the three points are you know non-collineal no observed in the point difference so what happens is that it's not going to be collinear guys 28 points 28 points sorry all the 28 points will not become uh non-collinear okay they'll be collinear points in that so we removed so how many will be that there you go seven c three eight into seven c three points foreign foreign foreign C3 and this is going to be done so I hope you got it guys [Music] foreign so I hope this is clear for all the students out there munchie questions given six horizontal and five vertical lines how many rectangles and squares will be made by their intersection okay any horizontal line is control five five vertical lines and any horizontal line is 4 4 Phi e and finally equal to 6. 6 horizontal lines five vertical lines asking you let us suppose each and every one unit is asking you how many uh rectangles and square roots according to the question by timings correct rectangle number all you have to do rectangle form of volunteer we could run the horizontal line you are going to Vertical is right now so all you have to do is six horizontal lines located here five horizont vertical lines will change one thing this will give you the total number of rectangles and from here we if there are if there are M horizontal lines M horizontal minus okay M horizontal lines and N vertical lines n vertical lines and the vertical lines if you want to find out in this particular case what is the number of rectangles well any particular scenario number of rectangles number of rectangles rectangles simply given by the formula m c 2 into nc2 in fact in only rectangles mc2 into nc2 so 6 c 2 into 5 C2 got it okay next next which is absorbs sorry foreign foreign [Music] one two three four five okay one two one two three four one two three four five so five into four five into four got it so one two three four and one two three four five okay so five into four is the first if only one unit squares all right two unit squats almost time simple one two okay and one two three and four so you put a simple Ranger star and a five into four comma T I could have four into three lasts [Music] you're going to who do and now go but in the complication in the Japan example symbols of finding the photos you just just stop it right at this point and this is going to give you the answer simple so is this clear for everyone out there so this is the concept of number of squares guys so five into four uh four into a condition is two three four five six seven five vertical lines this extend horizontal five vertical lines one two three four and two three four five value and either way like instead of taking them as lines okay so if you double the double to a shadow but it's fine like concept is the same logic foreign so I hope this is clear for all the students out there even a doubt on that I hope it's clear for you guys okay simple formula uh this is for rectangle actually chess board is chess board chess board this is right it will have eight units eight cross eight and eight chessboard which is what problem is so we get one two three four five six one two three four five six seven eight one two three four five six seven and eight okay so one two three four five six seven eight three six eight one two three four five six seven eight minutes [Music] number of rectangles number of number of rectangles the number of rectangles how do you calculate it I told you any vertical lines and horizontal lines so one two three four five six seven eight nine one two three four five six seven eight nine so nine c two into nine seed is going to give you what the number of rectangles okay and number of ID what is the number of this one guys the number of squares so number of squares key we started from eight cross eight and then we go to seven plus seven the one unit Square seven two units Square seven cross seven units five six plus six so continue also it will go to one cross one over so this simply nothing but summation eight Square okay so nine c two into nine c two and the conjugate and nine C2 into 9 C to calculate so nine factorial by two factorial into seven factorial so enter the unit of circular so nine into that's gonna become eight 36 into 36 60 36 63s are 18 18 that's going to be 36 39 degrees are 9 2.296 one two nine six is whatever I get one two nine six is what you are going to get the number of rectangles one two nine six one two minus six okay right and eleven thousand different summation eight Square random summation 8 square the summation n square and running from 1 to 80 which is n into so 8 into 8 plus 1 9 into 12 plus one two eights are sixteen plus one Seventeen by six single so two threes are two four right three ones are three things so three fours are twelve right so Seventeen into twelve in the 17 into 12 two sevens are fourteen seven plus two nine ten and this is one one and this is two so two not four so it's going to be how much 204 204 number of squares 204 and if the question is number of rectangles which are not squares according yes sir one two nine six minus two not four J and D you get down so so I hope this is clear for you so this is how you have to calculate guys so chessboard problem chess board of eight cross eight uh so approaches you start from the small smallest Square possible until one unit squares two units okay so I hope this is Cleo for everyone out there so now what we will do is let's go to the next one and this is basically now that we are completely equipped with NCR let's do some example using selection and Arrangement under entry selections for example 10 books online how many ways you can do this how many ways we can select three boys and two girls out of this money and arrange them on five six hello right first is two girls four feet okay and arrange them on five seats in how many ways can be arranged in five factorial yes or no it's like our distinct objects me are places three distinct boys and two girls five factorial foreign the total number of ways of selecting the level layers 13c9 is going to give me the total number of ways of selecting the level things 11 players took a batting order under ND 11 players the first second ever both so how many ways can be decided 11 factors simple so is this clear for you guys simple question is the number of functions F from 1 2 3 so 120 on to this particular this one Center such that F of K is a multiple of three whenever K is a multiple of 400 questions it might seem tough and the tough question really like the first cycle one two three so one one two three four so on 21 kitchen one two three so on reload up 21 kitchen okay now what are they saying is that you don't need much a low function you don't do such module a function define okay a function yellow Define whenever K is a multiple of 4 K is this person K is a multiple of 4 F of K is a multiple of three what does this mean guys twenty or spine and F of K multiple of 300 F of 3 6 9 12 15 and 18 50 and 80. particular Phi number is low a number this is that is what it is okay so how do you do this one guys very very simple okay dangerous Tower it's like this four digits low until you imagine like this question is four places slow 4 8 12 16 20 low three six nine twelve fifteen eighteen you can basically place any of this it is going to get mapped the same goes with eight the same goes with 12. the same goes with 16 the same goes with 20 also right so in this particular case how are you gonna do it in this case and he's saying that number of functions F from the set on to this set such that F of K is a multiple of three whenever K is a multiple of 4 correct okay so we can say that definitely foreign is foreign so I hope this is clear for all the students out there we go into function idea I think this question is not a big deal on to functionality function chapters so now let me go to another question in how many ways seven boys and four girls can be arranged in a row such that three questions which are in the row let me start doing the first question suggestions we do in this chapter Clarity we understand what is happening so let us see this one now how many ways seven boys and four girls can be arranged their row such that no two girls are together simple above try to understand in this way uh seven boys and now four girls okay uh no two girls are together no two girls are together boys will be together correct the first thing is foreign foreign foreign NPR is nothing but R factorial into NCR okay fine so now let me go to the next one all girls are together foreign foreign [Music] foreign okay so now let's go to the next one not all girls are together this is [Music] total minus all girls are together total minus all girls together so what does that means uh how many ways can you arrange them seven plus four eleven factorial minus of all girls are together eight factorial into four factorial that's going to be the answer okay so I hope this is clear for everyone out there so Jee made 2020 question two families with three members each and one family with four members are to be selected two families online let's say family number one and family number two with three members each and one family with four members are to be selected in a row okay in how many ways can they be seated so that the same family members are not separated and then the family members first family second family third family a family members first family three remember three members foreign foreign [Music] [Music] so out of this can be arranged foreign okay until let's say that foreign let us suppose five six objects are distinct number six same or distinct objects right foreign it's not going to be 10 factorial it's going to be 10 factorial by a by the objects which is 4 factorial in this particular case objects so let us say p objects out of that if P are P objects are alike P object storage let us say that foreign type of word which we can form using those letters you are assuming that all these letters are distinct but that is not the case foreign certain number of objects this is how you have to do it okay so for example let me take questions for you uh for example one two three four five five factors so buy two factorial into two factors the same goes with this also C and D so I hope this is clear for everyone okay so now guys what is the question given for you how many different nine digit numbers can be formed from the number two two three three five five three eight by rearranging its digit so that all digit occupy even positions or digits occupy even positions how many different nine digit numbers so one two three four five six seven eight nine okay digit so that all digits changes and even uh positions occupy just so this is an even position even position the low or digits probably okay and the military so what are the odd digits and even digits three three five five these are the odd digits two two three eight are the event so first even slope odds foreign factorial three two times repeat after me five two times two times so two factorial eight times only three times so eight sorry three factorial so is this clear for everyone out there I'm gonna doubt on the equation I hope it's clear for you okay that is what is given for you guys here so let me go to another one Jee made 2020. total number of 6 digit numbers in which only and all the five digits have to appear only if I use clearly one two three four five six seven foreign clear okay now one three five seven nine e five positions [Music] okay so what is what happened so this will be having six choices this will be having six choices okay save the five into four into three into two into one of the only six factorial options six numbers I think it's five days of communication it is going to become five days low so number of six digit numbers okay in which only and all the five digits one three five seven so five four three two one it is clear okay so this can be filled in five ways okay so what am I going to get five factorial into five divided by two conditions oh so okay and six factorial by two okay okay okay okay guys I'll make it better for you right there is more better for you guys bro one two three four five six seven okay uh this one a the E5 position six positions locally five positions what is going to happen is 13579 what is going to happen so basically position foreign foreign foreign [Music] is anyone having any any doubt in this particular case is this clear for everyone right okay so now let us do one thing another question in fact another concept um important conception iconic okay funny important special type of problems of time and those are this type of problems a b the arrangement grid problem dictionary problem sum of devices and exponents of prime p in a factorial these questions let us go to de-arrangement B Arrangement is summation of formula rearrangement concept anticipant for a suppose let us say okay okay you find the arrangements of this letter say more letters this is such that such that no letter will remain in its original position is the number of Arrangements in which original positions [Music] foreign [Music] okay so that is the concept of the arrangement so dearrangement I'll just write it down for you so that you know you guys can refer it afterwards the number of arrangements number of number of Arrangements the number of Arrangements in which in which in which the object the object the objects does not remain in its original position does not remain does not remain in its original position let's original position so either not a concept of rearrangement so depending upon the number of letters given for you you know okay let us say 3 e n factorial into 1 minus one by two one by one factorial in the formula okay which is which is basically calculating the dearrangement NHS a number of objects given for here argument okay so usual guys and letters foreign the number of ways such that no letter is dropped in its corresponding letter box okay 10 letters calculate the number of ways in that case so basically they are calculating the D arrangement s according to 10 factorial into a minus 1 by 1 factorial plus one by two factorial minus one by three factorial plus one by four factorial minus one by five factorial plus 1 by 6 factorial minus one by seven factorial plus one by eight factorial minus one by nine factorial plus 1 by 10 factorial okay but tenants usually give it till four or five okay but this is how you have to apply this so I hope it's clear so formula clearance I'll take a question for you for example let's say find the number of ways number of de-arrangements of four letters okay how many ways are such that only one letter goes to its original questions find the number of de-arrangement of four letters and changes so simple D four yes now which is 4 factorial into one minus one by one factorial plus one by two factorial minus 1 by 3 factorial okay so minus of 1 by 3 factorial is 1 by 4 factorial the rearrangement of four letters at the second question if four letters locally letter L will go to the own address so what does that means you need to select you need to select which letter will go to its own address so four letters locally in how many ways you can select that one you know letter which is going to its own address for Sigma foreign [Music] okay so fine I hope this is clear for you let me go to the next question next product next type of special problems right now great problems great problems find the number of short test parts from A to B number of shortest parts foreign foreign first question from A to B remember first stage is current day horizontal uh horizontal limited any units vertical units let us suppose if there are M horizontal units so from A to B let us say m horizontal steps I think steps is a better word horizontal steps and N vertical steps in the encounty and the vertical steps one horizontal two horizontal three horizontal step one two three vertical steps so n vertical steps on my unconding the number of shortest parts number of shortest parts formula and others are the number of shortest paths which is M plus and whole factorial n plus n whole factorial divided by m factorial K to n factorial this is going to be the formula okay so this is the formula so basically if all you have to do is calculate M and N guys so M values m is nothing but what horizontal steps so one two three four five six seven so Ms seven are n Anthony vertical sensor one two three four five substituent value 12 factorial by seven factorial into five factorial is going to be the answer so I hope this is clear for all the students out there okay from a to b y r c is horizontal structure one two two over horizontal steps vertical steps online so it's 5 factorial by two factorial into three factorial and C to B numerator sorry horizontal steps which is going to be seven factorial by factorial it is 7 factorial by 5 factorial into two factorial this is going to be the answer okay so I hope this is clear for everyone out there So This Is The Answer simple my J is a vertical horizontal okay so I hope this is clear for you so now let me go to another type of problem which is dictionary foreign made up of all words five letter words of all the words five letter words that can be formed using surats each letter used only once okay so dictionary loss that's where I have done my engineering nit Surat mechanical super place that was my best four years I've spent okay beautiful college a beautiful atmosphere minimum one two kilometers the reason the reason you know exams kinta competition the colleges which are offered the premier colleges India Love one of the best colleges of facilities exposure you know me me me thinking okay but it's not to just say that is give your best hundred percent is foreign [Music] Surat they need questions foreign [Music] generally is foreign what is going to happen you get R next so s r one two three window three factors in the s u is foreign 48 54 60 66 68 Surat is going to be the 69th word 69th word okay uh dictionary so I hope this is clear for everyone out there so a concept to this question is n factorial next concept is this so let us start discussing this in detail is this answer okay foreign foreign of prime p in N factorial for example questions okay find the number 2 raised to 4 into 3 square into 5 Cube you know for this number find the number of divisors foreign factorization design okay so once you have done the prime factorization what you can do is uh just a minute guys thank you okay uh right so prime factorization so let us suppose that you have this n okay they need prime factorization Json okay so prime factorization this is what you're gonna get in prime factorization so basically uh let us suppose 2 raised to a 3 raised to B 5 raise to c 7 raised to D so basically you'll get such type of you know prime factorization try to understand if you want to find out the number of divisors in this particular case the number of devices in this case it's going to be given by the formula a plus 1 plus 1 plus 1 B plus 1 so um okay foreign

hello everyone welcome back to your favorite Channel this is Gautam me mathematics Master teacher and I&#39;m back with another exciting session so guys which is permutations and combinations let us start this session today okay so first permutations and combinations [Music] permutations and combinations and combinations and combinations foreign different type of problems a good number of let&#39;s say Okay 20 or minimum 20 30 problems yes I session look a problem what is the approach I&#39;m following for the questions permutations and combination permutations and combinations permutations and combinations foreign we are going to discuss because six topics fundamental principle of counting very special problems like the arrangement problems and distribution of objects so six topics with the usual the questions a competitive examination E6 topics questions so let us understand in a detailed way permutation and combination so first guys which is fundamental principle of counting low fundamental principle of multiplication so for example let us say let us say foreign [Music] let us suppose foreign [Music] foreign okay this is like kind of real life scenario right from your general College foreign [Music] foreign [Music] of course foreign foreign foreign foreign I can give you another example let us saying for example let us say for example is okay let us say uh Hyderabad foreign [Music] foreign [Music] foreign foreign and the answer is again going to become 6 so this is the fundamental principle of counting low multiplication concept and a b multiplied same thing is given for you okay now let us quickly do a few questions practice key okay funny questions number of parts from A to B are four from B to C are three then find the number of paths three three a to c why are B how many number of B&#39;s you&#39;re gonna get in this case 4 into 3 and the answer is simple foreign [Music] foreign foreign foreign [Music] foreign multiply 7 into 6 into 5 is going to give you the answer so 6 5 30 into 7 210 there you go we answered so I hope this question is clear for you guys okay in fact questions which is fundamental principle of addiction multiplication let us suppose let us suppose um again of three options foreign foreign foreign foreign room foreign can you tell me any number of ways low foreign foreign foreign foreign so I hope it&#39;s clear for you guys okay foreign foreign okay so I&#39;m just telling you formula major like theorem is generally so let us see suppose you know which I do find the number of four letter words with or without meaning which can be formed out of the letters of the word make with the repetition of the letters is not allowed image which can be formed out of the letters of the word make foreign [Music] let us suppose first place first place [Music] foreign foreign of course one day each and every place name a four letter word law each and every letter any number of ways of course you have to multiply this so four into three into two four threes are 12 12 those are twenty four so total number of ways [Music] Common Sense okay so now let me do one foreign [Music] such that the vowels are in extreme positions extreme positions first and last position you know the first position of e particular position though and E particular positions simple first you okay so basically a particular position any waste of LGH so that is two weeks okay and you get a repetition not allowed to get a question you guys try to understand this foreign foreign [Music] so I hope this is clear for you guys don&#39;t get confused if you do four into three into two into one here it has four factorial of course right the factorials product of first and natural numbers slope n is always going to belong to the positive integers including 0. okay it is going to belong to positive integers until natural numbers including 0 and remember zero factorial value which is only okay so I hope this is clear for you later the month okay how many vowels sorry how many words can be formed rearranging the letters of this Delhi such type of woman&#39;s occupy even places so basically Delhi low okay you want to do um how many words we can get such that vowels occupy even places even places so this is an even place this is so that is and you have to fill this tree this is very simple three ways we need two ways so what is the answer sorry 2 into so basically multiply since you need to find out the entire word so 2 into 3 into 2 into 1 the last question three factorial so two into three factorial is going to be that sure so option A is going to be the right answer for this question so I hope it is so now let me go to another question for the question how many four letter words four letter words containing G can be formed using the letters of daughter and in order conditions repetition not allowed reputation a lot so a daughter only let us use four letter word forms only foreign [Music] G has four options foreign foreign foreign foreign [Music] foreign guys [Music] foreign foreign according to the question [Music] I&#39;ll remove all the words in which G is not contained the total number of words flow any words of words how many number of words do we get in which G is not there very simple only four letter words the total number of words in which G is not present and ultimately 7 raised to four seven raised to 4 8 raised to four which is number of words in which G is not contained and things are uncondi whatever the remaining result you&#39;re gonna get words so I hope this question is clear for you guys so reputation allowed first reputation questions okay so I hope this is clear for everyone let&#39;s do a question how many three letter words can be made using the letters of the word daughter such that if the word contains G then it must contain H also foreign G is selected so Jesus is [Music] two ways hello you can say it can be filled in two ways so G the two ways actually it&#39;s not two ways G converter okay so let us suppose foreign foreign [Music] six [Music] six twelve eighteen twenty four Thirty and thirty six it was clear for you guys six six six six different number of busy one 6 plus 6 12 and even in different cases evenly okay you should not be multiplying it is maybe so six plus six twelve is six eighteen twenty four Thirty thirty six or two minutes hours G not a selected foreign [Music] foreign [Music] 246 let us go to some problems or numbers [Music] and combination chapter law numbers where the questions of style like certain digits will start will say that a certain digit store D five digit number or four digit form number forms a such that it is divisible by 3 divisible by 2 divisible by five different different ways foreign [Music] how many four digit not possible repetition of digits can be made with no digit less than five basically by four digit number such that digit liquids and digits can be repeated okay but lower digit should be less than no digit should be less than five so what does that means anything zero one two three and four e digits use shares Colony you need to basically form a four digit number forms place so this can be filled once placed foreign [Music] foreign foreign 125 into 4 that is 500 so I hope this is clear for everyone out there okay with every digit with every digit less than 5. okay but you get the logic right I no digits eight and nine of this one six seven eight nine okay so I hope this is clear for everyone in fact so digits more than five and a particularly six is seven eight and nine okay so I hope this is clear for everyone out there okay so now guys sorry how many four digit numbers can be formed using one two three using one two three how many four digit numbers can be formed use JC four digit numbers definitely the repetition is allowed okay so we can say this can be filled in three bits can be filled in three three ways and three days so the answer should be 3 raised to four simple okay so nothing complicated in this particular question next question given the digits one two three four foreign how many three digit numbers less than 500 can be formed if each digit is used at most once second how many of those are divisible by five okay the first place observed any question the first question is one two three four five nine number three digit number forms how many three digit numbers less than less than 500 can be formed if each digit is used at most once okay okay guys so you can take only this number so this position can be filled in four weeks foreign foreign the right answer for this question is going to become 80. so I hope it&#39;s legal for you first place one two three foreign how many of these are divisible by five how many of these are there one two three and four and a four ways using those digits only foreign foreign foreign going to become so is this clear for everyone out there simple previous question learning first position is final positions which is going to be 16 okay so I hope this is clear for you in fact okay how many three digit numbers less than 600 and divisible by 5 can be formed using 0 1 2 3 4 5 9.3 because less than 600 so I hope you guys can do this strategy question okay so now let me go to another question it already changes let&#39;s go to this question guys how many four digit numbers can be formed using one two three four which are divisible by four which are divisible by four divided by four antioxide what is the divisibility test for four guys divisibility test for 480 last two digits should be divisible by four column so until it is so you&#39;ve been fixed value 24 2 square 24 32 this question 32 this version so one two three twelve uh 24 32 until so basically places you know three ways of three ways 12 24 32 foreign guys [Music] uh here the repetition is allowed how many four digit numbers can be formed using one two three four which are divisible by which are divisible by four so e k slow we can take reputation is allowed because I mentioned the other question right reputation I lovedi then we can say that this can be filled in four ways right this can also be filled in four ways okay so 4 4 16 and last number 241 21 23 24 uh 41 okay 31 32 33 34 41 42 43 44 perfect so what is the difference maybe I might be making some mistake in this case okay Mission guys edit 12 . sorry sorry I think uh I get it application sorry blue anime so let&#39;s suppose he named 12th of illness is low if you fill the last two digits by 12. let us suppose which one with 6 12 30 to 24th this is how you get it adjourned up sorry 12 24 32 and half 44 foreign [Music] right so if you do it in that way then clearly see that means four ways to repeat GH 0 and the remaining two places can be taken in four into four which is going to be uh definitely uh 16 ways the 16 into 4 64 answer key right so options 64 in the 64 option level 125 yeah sorry sorry I think there might be something uh wrong options so answer is going to become 64. sixteen fours are it&#39;s going to become 64 not 125 the right answer for this question is 64. okay so simply right I hope this is clear for you guys so for divisibility test tells in the last two digits so now let&#39;s go to another question in fact a question let me tell you one important point a point question is actually uh let us say you have to form a three digit number under 2 which is either divisible by 2 or 300 either divisible by two or three same question divisible by two or three energy so a lot go with this formula this and this is n of B minus n of a intersection sorry so equation you want to look sorry how many four digit numbers can be formed from which either begin with 3 or end with 8. how many four digit numbers so question number is how many four digit numbers can be formed which either begin with a or end with b a a begin with the 3 or end with eight I think let us suppose I&#39;ll consider this a as the number of ways in which this particular number okay is going to start with 3m so a or J C the total number of ways foreign how many how many four digit numbers can be formed which either begin with three or end with eight okay so in your sorrow we make it up three places slow or three or less three places [Music] the first and last position to 0 3 and 8th of agent a intersection b and a and t first three last eight under three guys got it is [Music] right so I hope this is here for you this is how you&#39;re gonna get the answer okay so that is so I needed this quantity so this can be found in 10 ways 10 waves and 10 ways so enough a 10 Cube also okay okay so this can be filled in 10 ways in 10 digits firstly 10 ways to fill J alone Square nine into ten Square and what is n of a intersection B another intersections so run the place in the middle foreign foreign what did we take a as a and a in this start with three but a complementary no it doesn&#39;t start with three so B and 10 is going to start ends with eight so we complemented not ending with it so a complement intersection B complement which is n of a union B whole complement the D Morgan is nothing but total the total of chasing you already know this n of a union B value substitute Json B you are going to get done so so in the equation under e so this is how you have to do it guys simple question either divisible by this or this is foreign s which are you know very helpful for us in counting year two results the number of ways of arranging are distinct objects at our places the number of ways of arranging are distinct objects that are clicks four places four objects online for four ways foreign foreign okay in fact we can also discuss the concept of permutation guys permutation is angular Arrangements foreign [Music] foreign [Music] foreign so I hope this is clear for all the students out there okay fine let us get back to this one if you don&#39;t talk ceremony number of ways of selecting you grow Arrangement typing Arrangement key selection key difference only you know order important let us say Okay um order report foreign 40 members low foreign so basically selection the order is not going to be important so remember that so okay let us suppose n objects and distinct objects and distinct objects are number of objects the formula is n factorial by R factorial into n minus our whole factorial in the formula right in fact foreign I think I mean the case of arrangement the order is going to matter of the way order matter of something first ever second level third episode [Music] order then long in the case of the order that&#39;s not right order which is an order does not matter in the case of which one in the case of selection so e Arrangements we are calling it as permutation the arrangements permutation and the selection on the combination permutations and combinations permutations and combinations basically Arrangements so I hope this is clear for everyone out there okay foreign foreign thank you foreign foreign foreign [Music] foreign [Music] you can start with selecting four men first case Okay selecting four men four men of course only one woman five members committee is one woman so how many ways can we do this so basically six men low four months anyway and five women select German H1 then three women three women so three women selectors so three women and two men okay so three women from five womens five c three and two men Allah from six men six feet so these are the two different so what is going to be the answer is if you add this whatever the result you&#39;re gonna get that is going to be the answer so got it simple question so four minutes containing four men and three women 64 into five three five guys I think uh the question below first question low okay I&#39;ll tell you the difference by a committee of five members five members foreign so I hope this is Cleo for every one of the requesting Theory this is going to be the answer okay this is what you want to get okay and coming to the next question how many committees contain at most two women at most two men sorry women so let us suppose you get a question within five members okay restriction Zero menu and five moment of Star Trek that is one scenario or else one man and four women of Sanitation or else Two Men and three women who started committee right and we can&#39;t go more than two because at most two men in order so what is how many ways can we select this on how many ways can selection be done in this case so 0. has to be selected which is 5 C 4 and this one two minutes so 62 into three women select Journey five women windshield it&#39;s going to be five c three which is to end up getting the clear I hope this is clear for you simple question is you have to get the answer as 181 okay so fine now let us go to one thing let us do one thing guys let us go to the next question how many committees contain at least one woman at least one woman one woman and four men the five members only so one woman four men presented two women Three Men and three women two men isn&#39;t a four women one man listen to five women so this in this many ways I can select a committee so the first one over men&#39;s energy so 64 into 5c one lady next three men sixty three into five C two will be foreign so I hope this question is clear for all the students out there depending upon the criteria given for you in the question so I hope this is clear for you so now let me go to another question a group of students consisting of four boys and five girls find the number of ways of selecting a team of at least three boys and four girls can you guys do it find the number of ways of selecting a team of at least three boys and four girls so at least a group of friends consists of four boys and five girls okay find the number of ways of selecting a team of at least three boys and four words so at least three boys boys and girls slow minimum three so how many ways in how many ways can you select three boys and four girls together so it&#39;s going to be four C three spicy forever next you can also go with four and five so minimum at least one day you can also go with four boys as well as five girls or do you know four boys and four girls at least three boys and four girls four boys four girls could have corrected in that case you get four C four into five C four or you can set three boys and five girls yes that is also Pro characters at least three at least four so this is like four C three into five C five five simple and finally you can go with four points and five months four points and finals and that is four C four into five C five just add them away whatever result you are gonna get that is going to be the answer okay so I hope this is clear for everyone only simple question is you can easily understand how to do the selection in this particular case you can see here the same thing one two three and the fourth case okay so now let me go to the next one Jee main 2020 question questions there are three sections in a question paper okay so let us see that section 1 Section 2 section three okay and each section contains five questions okay section one of five questions is section two or five questions so section three a lot of five questions are fine a candidate has to answer a total of five questions choosing at least one question from each section okay in the number of ways in which a candidate can choose the questions at least one uh questions foreign try to understand this I can do it in this way see this is about the number of ways you can pick up here so I can say that okay okay um what we will do is uh let&#39;s decide let&#39;s decide think about it is so that is going to be in how many number of ways three ways are the three ways so now you can say five c one five c one into five c three this is one way plus so that is why I&#39;m telling you again this is in three weeks into monthly five c one into Phi C2 into Phi CO2 so if you do this you are going to get done so simplify J and D children that is going to give you the answer so you can see here clearly same thing 3 into 5 c 1 5 c 1 5 C 3 3 into 5 C one five six two five C two simplify this are unconding the answer is 2 to 5. so I hope this question is clear for everyone okay so first changes foreign blue the kind of cricket team ongoing a cricket team low okay players so how many ways we can select it 14 C11 so I got it everywhere so a particular player has never chosen Nikita players so 14c 11 is going to be the answer for this okay next question or not sorry three particular players are always chosen 15 players foreign foreign and eight players knee is and that is the number of ways in this particular case so got it everyone so simple I think I think is also clear for eign there are two ways of doing questions total number of ways in which you can select the team and I&#39;ll select jh2 simple you have 15 to 3 11 select General 15 C 11 select JS number one okay 15 C11 in this case also what happens is that there are number of ways the num there is number of ways in which two particular players these two players should never come together I&#39;ll remove all the cases of all the Selections in which I have selected both the players okay in how many ways can we do that guys simple the first angels are entered two players are going to come together 15 C 11 noon shall remove all the ways all the Selections in which these two players have come together 15 C 11 minus our under players okay the two particular players playing together minus two particular players together get the number of ways in which they are not together right you get the number of ways in which they are not two together so total 15 is 11 two players come together and there are two gates yes is going to be the answer so I hope this is clear for everyone okay same thing but both you end up getting the same answer guys okay so I hope it&#39;s clear for you for you so now let me do this question guys in how many ways we can select D numbers out of one two so 100 such that they form a a b they form and they are basically going to be in arithmetic combination so basically let us suppose X1 X2 and X3 these condom numbers okay such that they are they have to be in AP Allah so always just select jh123 234-345-456-678-13579 okay okay so since they are in EP what we can write is 2 x 2 is going to become X1 plus x 3 right when three numbers are in AP this is the criteria or I can say X2 is going to become x 1 plus x 3 divided by this what is clear is X1 plus X3 should be divisible by 2 yes or no X1 plus X3 is it the number because this should definitely be divisible X2 is going to be a natural number of product possible X1 plus X3 should be divisible by so how can we choose X1 and X3 so sum of two numbers divisible by 2 1 day the number should be even what is the divisibility divisibility test of two the number should be even so we know that this X One Plus X3 so we know that x 1 plus x 3 is a even number it&#39;s going to be an even number either sum of two numbers even I put out on the there are two cases for that either x one and X2 are both even you don&#39;t do even numbers we address the even number of something x 1 and X2 both are odd numbers I&#39;m going to odd numbers you add Json odd numbers odd numbers zero you get the result as even so now you calculated how many ways you can select even numbers two even numbers out of these numbers and how many ways you can select odd numbers out of the given numbers so let us suppose so for this one so for this one for this one so any even numbers only one to hundred modulo 50. so 50 even though you don&#39;t want to select generally so how many ways okay similarly it&#39;s for an extra run to odd any odd numbers one time only 50 so 50 kilometers so 50 c two foreign [Music] so I hope this is clear for all the students out there so 2 into 50 c 2 is going to be the answer okay so munchie question will be conceptually good question so I hope you guys understood this one is [Music] here so this is can be done permission guys so this can be so let us say in the first case second case so case number one 50 c two ways case number two fifty C2 base loss this is a total interest on the graph it&#39;s going to become 2 into 50 50 c two so I hope this is clear okay so now let me go to another question in fact only method discussions okay that is okay so now let me go to this concept guys the total number of selections out of n distinct objects is 2 . and distinct objects one two three four so on and so one two three four so on and distinct objects okay so total number of selections until total number of any ways to select JH objects out of and desktop sorry out of n distinct object flow out of and distinct objects law any way slow objectives foreign foreign foreign foreign you know foreign is going to become so I hope this is clear for everyone okay so guys let&#39;s go to the questions there are three distinct physics books that four distinct mathematics Books Okay in how many ways books can be selected and the ABC questions in Shadow question number is foreign foreign services the total number of ways just add it up the 7c 0 plus 7 say one plus seven C two plus seven C seven which is nothing but sure the answer clear oh the question number eight let&#39;s go to question number B at least one book is selected at least one book is selected physics foreign [Music] foreign so I hope it&#39;s clear for you guys okay next question is selection can be made such that at least one math book is selected at least put a specifically um how many number of ways I can select the books simple okay at least one books so foreign foreign special case of selection is foreign foreign foreign in how many ways can I pick up those three chocolates out of 10 identical chocolates one way one way okay okay so from this concept which you have to understand is La n identical objects in identical objects n identical objects and identical objects okay the number of ways in which you can pick up let us say one object two objects anyways three objects anyways so on um the number of ways of selecting two A&#39;s out of Four A&#39;s is what [Music] selecting let us say zero or more objects selecting is low n identical and identical objects in identical and identical objects in identical objects in G if you want to select okay if you want to select zero or more objects select 0 or more objects here or more objects so zero or more objects this is nothing but n plus one the number of a is n plus one identical object slow object selected selection n number of times it is n foreign a foreign foreign three letters selection I get three different letters three different letters three letters different guys what are the different cases three letters and ten let us say more distinct number three is foreign next two two distinctive sorry two I like one distinction one distinct number okay so how can we how many number of faces in the world and I could have two allocated two letters foreign [Music] foreign [Music] this is going to become and finally three alike three are like anyways so this can be selected in two weeks so the total number of ways of selecting three letters six plus one three three plus two plus six plus one three guys I&#39;m very sorry six plus one seven seven plus two nine so therefore nine ways is going to be the total of every particular case foreign foreign let&#39;s go to the next step concept and that is this question and the equation law in the application in how many ways we can select four letters out of assassination assassinations okay okay okay foreign similar to the previous question four letters four letters four distinct this question first case for a distinct any distinct letters is also four so anyways to the H is [Music] educational 64 and that is this one for distinct so four distance do I like and two distinct to alike and two distinct so two alike and that is so run the same letters you can take either two A&#39;s or two s&#39;s or two eyes or two ends four options foreign three alike wonderful three alike and one District level rights one three I like and one distinct and Rascals how can we write this down there is let us suppose a select j7 and finally 4 alike four same letters sorry only one way and the content only essay foreign you get done so if you add up all these cases you get the answer so arthma in the clear Gap I hope it&#39;s clear for you then so first thing is 64 right which is four C one into five C two four C one into five sorry four C two and I can two District Four C one one two three four five foreign and two I like to alike and type 2 una four letters will be like a a s s there is one two three four foreign for everyone out there so another two alike and two alike and three are like at one desk and already choose something it&#39;s two c one into five c one save money two c one into five c one and finally all I like is only one so if you add it up you get 72 as an answer okay so this is the question munchie questions so I hope you get it okay fine so now what we will do is let&#39;s see some geometrical counting problems counting problems selections and a combinations the first question for you guys there are end points on a circle circle final end points online first question entry joining this points how many lines will be made how many lines will be made okay so one two three four five six you have in there foreign since no point is I can write down since no point is scoring here no two points are calling here since no two points are convenient okay since no two points are convenient are cholinium what we can say in this particular case the answer the result is going to be in how many ways you can select the two points out of the given number of points that will give you the number of straight lines so this is the formula and C to another formula number of straight lines and C2 and every formula number of straight lengths when the points are given points straight lines that is okay next question in journal Theory joining this point how many triangles will be made how many ways you can select three points right since three points let&#39;s say suppose three points so since next one for you guys next one indicator joining those points of polygon is made find the number of its diagonals either number of diagonals concordance number of diagonals number of diagonals of a polygon of a polygon of enzymes of a polygon of n sites what is the formula nc2 minus l nc2 minus n or n into n minus 3 divided by 2 the formula number of diagonals concordance for n sided polygon okay so I hope this is clear for everyone out there okay so NC n into n minus 3 by 2 is going to be the formula where n is going to be the number of sides I think e particular question k n is number of points uh you do one thing guys let us say we simple the ionic equation is MC2 is going to become n e value of n substitute okay so I hope this is clear for all the students out there guys okay so this is how you have to do it uh right so save it Shadow for a diagonal which is the line we require two points but those selection will include sides of polygon also as they are all formed by selecting two points so number of diagonals of a n-sided polygon is nc2 minus n okay that the number of diagonals is n-sided polygon and nc2 minus N I hope it&#39;s clear for you okay so this is the formula covalented question ignore the same equation though equation question confusing the N points on the particular one suppose three sides questions I just cancel this question out foreign [Music] sides is going to be nc2 minus n right okay let&#39;s go to this question consider seven points four are collinear joining these points how many lines will be made first question triangles will be made you put a equation device joining this point how many lines will be made so users foreign foreign what is that formula is out of n points out of n points out of n points if M points are collinearium out of n points M points are collinear okay then then the number of straight lines the number of straight lines of a formula into simple gum and C2 minus mc2 plus 1 is a formula using that formula you can get the answer so I hope this is clear for you okay so next question is the next question is how many triangles have been made out of n points out of n points out of n points if M points are collinear okay M points are collinear then the number of then the number of which one the number of triangles the number of triangles simply nc3 minus nc3 nc3 points it will not form a triangle nc3 minus McRae a particular question answer 7c3 minus 4 CPM so I hope this is clear for everyone out there okay so now let me go to the next one text for example the question you want to talk sorry in a plane there are eight lights no two of which are parallel no three are concurrent okay how many triangles can be formed with their point of intersection as vertices foreign first number of point of intersections November of point of intersection at the point of intersections in allows in the two lines under two number of point of intersection is directly proportional and directly proportional number of intersects sorry the number of point of intersection is nothing but the number of ways of selecting two lines out of eight lines together we wrote eight lines since the lines are not parallel and no three lines are concurrent property so we can see that number of point of intersections is simply lc2 okay so what is the value so calculate the pseudometer so eight factorials by 2 factorial and 6 Factor here which is eight into seven y t so that went away 28 guys 28 number of point of intersection I think normally how many triangles can be formed under 28 points are number of triangles number of triangles number of triangles the directing are 28 C3 RS are 28 points will see three points can conserve such that you know three points three points anyways that is the number of triangles not only 28 C3 in the country 28 C3 low are you sure all the 28 points will be non-collinear foreign so any point of intersection for child three point of intersection three point of intersection which you will also get from the Formula 3 c 2 which is still I think it could absorb oxide what&#39;s the point of intersections any three are three can I say all the three points are you know non-collineal no observed in the point difference so what happens is that it&#39;s not going to be collinear guys 28 points 28 points sorry all the 28 points will not become uh non-collinear okay they&#39;ll be collinear points in that so we removed so how many will be that there you go seven c three eight into seven c three points foreign foreign foreign C3 and this is going to be done so I hope you got it guys [Music] foreign so I hope this is clear for all the students out there munchie questions given six horizontal and five vertical lines how many rectangles and squares will be made by their intersection okay any horizontal line is control five five vertical lines and any horizontal line is 4 4 Phi e and finally equal to 6. 6 horizontal lines five vertical lines asking you let us suppose each and every one unit is asking you how many uh rectangles and square roots according to the question by timings correct rectangle number all you have to do rectangle form of volunteer we could run the horizontal line you are going to Vertical is right now so all you have to do is six horizontal lines located here five horizont vertical lines will change one thing this will give you the total number of rectangles and from here we if there are if there are M horizontal lines M horizontal minus okay M horizontal lines and N vertical lines n vertical lines and the vertical lines if you want to find out in this particular case what is the number of rectangles well any particular scenario number of rectangles number of rectangles rectangles simply given by the formula m c 2 into nc2 in fact in only rectangles mc2 into nc2 so 6 c 2 into 5 C2 got it okay next next which is absorbs sorry foreign foreign [Music] one two three four five okay one two one two three four one two three four five so five into four five into four got it so one two three four and one two three four five okay so five into four is the first if only one unit squares all right two unit squats almost time simple one two okay and one two three and four so you put a simple Ranger star and a five into four comma T I could have four into three lasts [Music] you&#39;re going to who do and now go but in the complication in the Japan example symbols of finding the photos you just just stop it right at this point and this is going to give you the answer simple so is this clear for everyone out there so this is the concept of number of squares guys so five into four uh four into a condition is two three four five six seven five vertical lines this extend horizontal five vertical lines one two three four and two three four five value and either way like instead of taking them as lines okay so if you double the double to a shadow but it&#39;s fine like concept is the same logic foreign so I hope this is clear for all the students out there even a doubt on that I hope it&#39;s clear for you guys okay simple formula uh this is for rectangle actually chess board is chess board chess board this is right it will have eight units eight cross eight and eight chessboard which is what problem is so we get one two three four five six one two three four five six seven eight one two three four five six seven and eight okay so one two three four five six seven eight three six eight one two three four five six seven eight minutes [Music] number of rectangles number of number of rectangles the number of rectangles how do you calculate it I told you any vertical lines and horizontal lines so one two three four five six seven eight nine one two three four five six seven eight nine so nine c two into nine seed is going to give you what the number of rectangles okay and number of ID what is the number of this one guys the number of squares so number of squares key we started from eight cross eight and then we go to seven plus seven the one unit Square seven two units Square seven cross seven units five six plus six so continue also it will go to one cross one over so this simply nothing but summation eight Square okay so nine c two into nine c two and the conjugate and nine C2 into 9 C to calculate so nine factorial by two factorial into seven factorial so enter the unit of circular so nine into that&#39;s gonna become eight 36 into 36 60 36 63s are 18 18 that&#39;s going to be 36 39 degrees are 9 2.296 one two nine six is whatever I get one two nine six is what you are going to get the number of rectangles one two nine six one two minus six okay right and eleven thousand different summation eight Square random summation 8 square the summation n square and running from 1 to 80 which is n into so 8 into 8 plus 1 9 into 12 plus one two eights are sixteen plus one Seventeen by six single so two threes are two four right three ones are three things so three fours are twelve right so Seventeen into twelve in the 17 into 12 two sevens are fourteen seven plus two nine ten and this is one one and this is two so two not four so it&#39;s going to be how much 204 204 number of squares 204 and if the question is number of rectangles which are not squares according yes sir one two nine six minus two not four J and D you get down so so I hope this is clear for you so this is how you have to calculate guys so chessboard problem chess board of eight cross eight uh so approaches you start from the small smallest Square possible until one unit squares two units okay so I hope this is Cleo for everyone out there so now what we will do is let&#39;s go to the next one and this is basically now that we are completely equipped with NCR let&#39;s do some example using selection and Arrangement under entry selections for example 10 books online how many ways you can do this how many ways we can select three boys and two girls out of this money and arrange them on five six hello right first is two girls four feet okay and arrange them on five seats in how many ways can be arranged in five factorial yes or no it&#39;s like our distinct objects me are places three distinct boys and two girls five factorial foreign the total number of ways of selecting the level layers 13c9 is going to give me the total number of ways of selecting the level things 11 players took a batting order under ND 11 players the first second ever both so how many ways can be decided 11 factors simple so is this clear for you guys simple question is the number of functions F from 1 2 3 so 120 on to this particular this one Center such that F of K is a multiple of three whenever K is a multiple of 400 questions it might seem tough and the tough question really like the first cycle one two three so one one two three four so on 21 kitchen one two three so on reload up 21 kitchen okay now what are they saying is that you don&#39;t need much a low function you don&#39;t do such module a function define okay a function yellow Define whenever K is a multiple of 4 K is this person K is a multiple of 4 F of K is a multiple of three what does this mean guys twenty or spine and F of K multiple of 300 F of 3 6 9 12 15 and 18 50 and 80. particular Phi number is low a number this is that is what it is okay so how do you do this one guys very very simple okay dangerous Tower it&#39;s like this four digits low until you imagine like this question is four places slow 4 8 12 16 20 low three six nine twelve fifteen eighteen you can basically place any of this it is going to get mapped the same goes with eight the same goes with 12. the same goes with 16 the same goes with 20 also right so in this particular case how are you gonna do it in this case and he&#39;s saying that number of functions F from the set on to this set such that F of K is a multiple of three whenever K is a multiple of 4 correct okay so we can say that definitely foreign is foreign so I hope this is clear for all the students out there we go into function idea I think this question is not a big deal on to functionality function chapters so now let me go to another question in how many ways seven boys and four girls can be arranged in a row such that three questions which are in the row let me start doing the first question suggestions we do in this chapter Clarity we understand what is happening so let us see this one now how many ways seven boys and four girls can be arranged their row such that no two girls are together simple above try to understand in this way uh seven boys and now four girls okay uh no two girls are together no two girls are together boys will be together correct the first thing is foreign foreign foreign NPR is nothing but R factorial into NCR okay fine so now let me go to the next one all girls are together foreign foreign [Music] foreign okay so now let&#39;s go to the next one not all girls are together this is [Music] total minus all girls are together total minus all girls together so what does that means uh how many ways can you arrange them seven plus four eleven factorial minus of all girls are together eight factorial into four factorial that&#39;s going to be the answer okay so I hope this is clear for everyone out there so Jee made 2020 question two families with three members each and one family with four members are to be selected two families online let&#39;s say family number one and family number two with three members each and one family with four members are to be selected in a row okay in how many ways can they be seated so that the same family members are not separated and then the family members first family second family third family a family members first family three remember three members foreign foreign [Music] [Music] so out of this can be arranged foreign okay until let&#39;s say that foreign let us suppose five six objects are distinct number six same or distinct objects right foreign it&#39;s not going to be 10 factorial it&#39;s going to be 10 factorial by a by the objects which is 4 factorial in this particular case objects so let us say p objects out of that if P are P objects are alike P object storage let us say that foreign type of word which we can form using those letters you are assuming that all these letters are distinct but that is not the case foreign certain number of objects this is how you have to do it okay so for example let me take questions for you uh for example one two three four five five factors so buy two factorial into two factors the same goes with this also C and D so I hope this is clear for everyone okay so now guys what is the question given for you how many different nine digit numbers can be formed from the number two two three three five five three eight by rearranging its digit so that all digit occupy even positions or digits occupy even positions how many different nine digit numbers so one two three four five six seven eight nine okay digit so that all digits changes and even uh positions occupy just so this is an even position even position the low or digits probably okay and the military so what are the odd digits and even digits three three five five these are the odd digits two two three eight are the event so first even slope odds foreign factorial three two times repeat after me five two times two times so two factorial eight times only three times so eight sorry three factorial so is this clear for everyone out there I&#39;m gonna doubt on the equation I hope it&#39;s clear for you okay that is what is given for you guys here so let me go to another one Jee made 2020. total number of 6 digit numbers in which only and all the five digits have to appear only if I use clearly one two three four five six seven foreign clear okay now one three five seven nine e five positions [Music] okay so what is what happened so this will be having six choices this will be having six choices okay save the five into four into three into two into one of the only six factorial options six numbers I think it&#39;s five days of communication it is going to become five days low so number of six digit numbers okay in which only and all the five digits one three five seven so five four three two one it is clear okay so this can be filled in five ways okay so what am I going to get five factorial into five divided by two conditions oh so okay and six factorial by two okay okay okay okay guys I&#39;ll make it better for you right there is more better for you guys bro one two three four five six seven okay uh this one a the E5 position six positions locally five positions what is going to happen is 13579 what is going to happen so basically position foreign foreign foreign [Music] is anyone having any any doubt in this particular case is this clear for everyone right okay so now let us do one thing another question in fact another concept um important conception iconic okay funny important special type of problems of time and those are this type of problems a b the arrangement grid problem dictionary problem sum of devices and exponents of prime p in a factorial these questions let us go to de-arrangement B Arrangement is summation of formula rearrangement concept anticipant for a suppose let us say okay okay you find the arrangements of this letter say more letters this is such that such that no letter will remain in its original position is the number of Arrangements in which original positions [Music] foreign [Music] okay so that is the concept of the arrangement so dearrangement I&#39;ll just write it down for you so that you know you guys can refer it afterwards the number of arrangements number of number of Arrangements the number of Arrangements in which in which in which the object the object the objects does not remain in its original position does not remain does not remain in its original position let&#39;s original position so either not a concept of rearrangement so depending upon the number of letters given for you you know okay let us say 3 e n factorial into 1 minus one by two one by one factorial in the formula okay which is which is basically calculating the dearrangement NHS a number of objects given for here argument okay so usual guys and letters foreign the number of ways such that no letter is dropped in its corresponding letter box okay 10 letters calculate the number of ways in that case so basically they are calculating the D arrangement s according to 10 factorial into a minus 1 by 1 factorial plus one by two factorial minus one by three factorial plus one by four factorial minus one by five factorial plus 1 by 6 factorial minus one by seven factorial plus one by eight factorial minus one by nine factorial plus 1 by 10 factorial okay but tenants usually give it till four or five okay but this is how you have to apply this so I hope it&#39;s clear so formula clearance I&#39;ll take a question for you for example let&#39;s say find the number of ways number of de-arrangements of four letters okay how many ways are such that only one letter goes to its original questions find the number of de-arrangement of four letters and changes so simple D four yes now which is 4 factorial into one minus one by one factorial plus one by two factorial minus 1 by 3 factorial okay so minus of 1 by 3 factorial is 1 by 4 factorial the rearrangement of four letters at the second question if four letters locally letter L will go to the own address so what does that means you need to select you need to select which letter will go to its own address so four letters locally in how many ways you can select that one you know letter which is going to its own address for Sigma foreign [Music] okay so fine I hope this is clear for you let me go to the next question next product next type of special problems right now great problems great problems find the number of short test parts from A to B number of shortest parts foreign foreign first question from A to B remember first stage is current day horizontal uh horizontal limited any units vertical units let us suppose if there are M horizontal units so from A to B let us say m horizontal steps I think steps is a better word horizontal steps and N vertical steps in the encounty and the vertical steps one horizontal two horizontal three horizontal step one two three vertical steps so n vertical steps on my unconding the number of shortest parts number of shortest parts formula and others are the number of shortest paths which is M plus and whole factorial n plus n whole factorial divided by m factorial K to n factorial this is going to be the formula okay so this is the formula so basically if all you have to do is calculate M and N guys so M values m is nothing but what horizontal steps so one two three four five six seven so Ms seven are n Anthony vertical sensor one two three four five substituent value 12 factorial by seven factorial into five factorial is going to be the answer so I hope this is clear for all the students out there okay from a to b y r c is horizontal structure one two two over horizontal steps vertical steps online so it&#39;s 5 factorial by two factorial into three factorial and C to B numerator sorry horizontal steps which is going to be seven factorial by factorial it is 7 factorial by 5 factorial into two factorial this is going to be the answer okay so I hope this is clear for everyone out there So This Is The Answer simple my J is a vertical horizontal okay so I hope this is clear for you so now let me go to another type of problem which is dictionary foreign made up of all words five letter words of all the words five letter words that can be formed using surats each letter used only once okay so dictionary loss that&#39;s where I have done my engineering nit Surat mechanical super place that was my best four years I&#39;ve spent okay beautiful college a beautiful atmosphere minimum one two kilometers the reason the reason you know exams kinta competition the colleges which are offered the premier colleges India Love one of the best colleges of facilities exposure you know me me me thinking okay but it&#39;s not to just say that is give your best hundred percent is foreign [Music] Surat they need questions foreign [Music] generally is foreign what is going to happen you get R next so s r one two three window three factors in the s u is foreign 48 54 60 66 68 Surat is going to be the 69th word 69th word okay uh dictionary so I hope this is clear for everyone out there so a concept to this question is n factorial next concept is this so let us start discussing this in detail is this answer okay foreign foreign of prime p in N factorial for example questions okay find the number 2 raised to 4 into 3 square into 5 Cube you know for this number find the number of divisors foreign factorization design okay so once you have done the prime factorization what you can do is uh just a minute guys thank you okay uh right so prime factorization so let us suppose that you have this n okay they need prime factorization Json okay so prime factorization this is what you&#39;re gonna get in prime factorization so basically uh let us suppose 2 raised to a 3 raised to B 5 raise to c 7 raised to D so basically you&#39;ll get such type of you know prime factorization try to understand if you want to find out the number of divisors in this particular case the number of devices in this case it&#39;s going to be given by the formula a plus 1 plus 1 plus 1 B plus 1 so um okay foreign

Transcript for:Permutations and Combinations Lecture

Transcript for:
Permutations and Combinations Lecture