e e e e everyone how are you guys Hey sir I'm starting ADM do you think as level function is a good place to start un Sol same but uh I'm good alhamdulillah thank you for asking uh sir we cover the whole syllabus in these streams I'll try to uh I'll try to cover as much as possible but um I'll obviously prioritize the important topics so I'm good thank you for asking a so to answer your question yes as level math functions is a good place to start you'll at least understand uh the basics as in how domain and range Works s please Tri ex stream start I teach online entirely I am based in Karachi but I teach online entirely so whether I take it in Karachi or laor or Islamabad doesn't really make a difference a let me just add the name of the topic minute start just a minute Thursday by the way I'm streaming um binomial binomial theorem it's an easy topic I know but uh I'm basically leaving the important topics for the end okay here we are all right so let's start I hope sir what camera do I use I use um this camera called insta 360 link to yes so having difficulty any advice I'll make a video on it okay I'll make a video on it sir I'm Al from your adance P sir can I leak your number my number is pretty common anyway but yeah please don't do that anyway so let's start enough talk let's get to work so basically number Le I should probably put my phone on silent so permutations and combinations now we're going to cover the concept over here as well okay and we're going to try it and try and do it in as much detail as possible Alina you can still register okay you can still do that now first understand the code difference between permutations and combinations now when we talk about permutations we apply permutations when we have to select and arrange okay when we are we have to select and at the same time arrange something people say that permutations is used when you just have to arrange something that's true but remember you're not just arranging you're also selecting so you're first selecting and then you're arranging okay and this usually this can be applied to objects you can select objects and arrange them like you'll see questions uh where you're arranging Books Okay and uh you can apply this to people like for example if people are lining up for an interview you want to find out how many ways you can arrange them so you can do that you can do it to letters because if you change the order of the letters it becomes a different word that means order matters and you can also do it to numbers why because if you change the order of the numbers then what happens as a result is that it's a different number altogether so so we use permutation wherever we feel that if we change the order it will result in a different outcome so that is where we use permutation where the order is important okay where we feel the order is important that is where we use permutation now then let's talk about combinations now as far as combinations is concerned combinations is used when you are not concerned with the order because changing the order doesn't really make a difference okay changing the order doesn't really make a difference like for example if you're applying to college and the college says okay just just a random example so does that mean of course not I mean ultimately what they want is just two A's and one b in whatever order you get it they're not concerned with that you just need to have two a's and 1 B that's that's all that matters so permutation combination is applied when you are selecting only and this only happens in one case that is when you are making a team okay when you are making a team why because if you're making a team or if you're selecting people if the question categorically says selecting people in that case the order is not important actually I'm not going to read see I can I can only do one thing at once I can either read what you guys are saying or I can teach so I will pause in between and then take your questions so right now I would suggest that you hold it you'll probably find the answer to most of your questions along the way so AB is the same as ba why because it's the same people therefore it's the same team okay like for example someone says that you know I have an idea that let's put a and b together in a team but then someone else comes along and says you know what I have a better idea let's put BNA together and make a team well should the two start a debate over this of course not that would be a total waste of time because they're both saying the same thing it's the same two people regardless of the order in permutations however AB is not the same as ba okay ab and ba are considered as two different outcomes now for here are a few examples to help you understand when to use permutation when to use combination so if for example there is prize winning okay prizes have to be distributed and let's say here's judge one who says that you know according to me a should take the first position C should take the second position and B should take the third position and then there is some other judge who says you know what no according to me B should be first a should be second C should be third so are these two saying the same thing no they're not these two are not saying the same thing they have different suggestions the order is different therefore this is going to fall under permutation so because what judge one and judge two are saying is not the same therefore this scenario okay where changing the order like I said results in a different outcome falls under permutation so this is basically an example of permutation similarly if let's say phone password 2357 and someone says no please don't watch this for S1 S1 permutation combination playlist it's not complete right now but uh there are a lot of videos so I suggest you start watching that yes I will cover the whole chapter today inshallah with P papers there's one thing I'm going to skip that is doing NCR and NPR without a calculator for that I'll make a video separately anyway so someone says that you know the the correct password to let's say your phone is 2357 now can someone enter the password in any order and be able to unlock your phone of course not why because in situations like these when you're when you're arranging num num is changing the order results in a different outcome so this also is an example where it falls under permutation okay this is another example of a scenario that's under permutation okay now let's start with some basic questions let's say there are three people and you're supposed to arrange them in a straight line so how do you arrange three people in a straight line let's see let's do it manually if you know how to do it that's brilliant let's assume that at this point we don't know how to do it so we can say that we have a first and then B and then C so this is the obvious way of arranging them now let's see how many ways we can shuffle them so let's fix a and then let's C let's have C in the second place and then B and then let's see how many more ways we can shuffle them let's have B in the first place and then a and then C and then let's have B in the first place but Shuffle A and C and then we can do the same if we have C in the first place so here's what we can do duplicate so we can have C and then B and then a or we can have c a b is there any other possibility that you think I might have missed feel free to point it out but there isn't so altogether there are six unique ways that three people can be arranged in a straight line now does that mean we have to do it like this every single time of course not what if there are five people what if there are 10 people what are you going to do then so there is a shortcut the shortcut is that you look at how many options you have in the first place so if there are three people and you wish to arrange all three of them so that means in the first place we have three options in the second place we now have two options because one person doesn't matter whether it's a b or c but already arrange now we're left with two so that means for the second place we have two options and then for the third and the final place we just have one option so all together there will be six ways now either you can do this or you can do three factorial which I'll explain what 3 factorial is in a minute this is also equal to 6 or you can do 3 P3 okay so what's 3 P3 equal to 3p3 is equal to 6 now let's understand 3 P3 the number that we have before p is the total the number that we have after three is the number of people or the number of objects we have to arrange that's how it works okay so this was just a basic example we're going to do we're not going to do this every single time so this is basically we either do this or we do this so make sure that you understand both all right sir is permutations and combinations somehow similar to no no bubble sort and Mer sort they're two different things um you can use permutation I mean they're not alternates if you think they're alternates that's that's not true what do you mean by total total as in the number of people that we have and the number of people that we have to arrange so here we have three people and we have to arrange all three of them but in some cases you will notice that there are five people and you only have to arrange maybe three of them or four of them sir does permutation and combination in non-calculator paper it can it can come in paper one okay we don't have a lot of evidence as of yet you guys will probably be actually no Feb Mar students will be the first one to give the exam as per the new syllabus so we're going to find out then as per the specimen paper it was I think in a calculator paper but the non-calculator part the specific non-calculator part for this chapter I'm going to do that in a separate video okay uh bet that's irrelevant V that's irrelevant as far as admat is concerned that's irrelevant but just to answer your question we just subtract one from the total if you're arranging them in a circle okay now how many ways can five people be arranged in a straight line so we have five options and then four and then three and then two and then one so multiply it all together 5 into 4 that's 20 20 * 3 is 60 60 * 2 is 120 or you could have done five factorial or even 5 P5 they all mean the same thing which is equal to 120 okay now let's do Part B so Part B says how many ways can three or five people be arranged in a straight line so notice now we have to arrange not all five of them we have to arrange only three of the five people so that means in the first place we have five options in the second place we have four options and then in the third place we have three options so we're not going to go all the way to one just 5 4 3 that's good enough or what we can do is and this by the way is equals to 60 now what we could have done was we could have done 5 P3 which is also equal to 60 why 5 P3 because you have five is the total number of people that you have and of the five you only have to arrange three of them okay so this is another way that you could have arranged three people out of five now let's go over the definition factorial what's factorial factorial is basically the product of all positive integers less than or equal to n where N is a positive integer meaning if you have three factorial that would be equal to 3 into 2 into one if let's say you have five factorial so you'll start with five you'll keep on going in a descending order till you get to one so if let's say you have 9 factorial or 10 factorial that's going to be 9 into 8 into 7 all the numbers in between till you get to 3 into 2 into 1 that's what factorial is and what's what's the use of factorial the use of factorial is to arrange and different objects in a straight line so if you have 10 people and you wish to arrange all 10 of them in a straight line that would be 10 factorial if let's say you have 15 people you wish to arrange all 15 of them in a straight line that would be 15 factorial a uh one thing that I'd like to mention is that there are 37 people watching if you guys haven't subscribed to the Channel please do that and I would also appreciate if you like the stream all right now moving on Now One use case of permutation is basically arranging letters okay you'll find questions like these very common in past papers and they're very easy so let's see how things work how many ways can the letters of the following words can be arranged part A Ali there are three letters all three of them are unique so the answer would be simply three into so for the first place we have three options for the second place we have two options for the last place we have one so it's 3 into 2 into 1 or simply three factorial which is equal to six okay then we have Bob okay how many ways can you arrange Bob you were helpful for me and R I appreciate it uh what's that yes I will be doing calculus but not right now obviously later on inshah let's arrange Bob and let's do it manually uh no the answer is is not two let's find out so let's first start with the most obvious one and that is ball now how about someone comes along and says you know what let's swap the bees will it make a difference I don't know you guys tell me does it make a difference then another way would be the following BB o and then someone else comes along that and says you know what let's swap the B and we have another Arrangement okay let's see if that makes sense and then the third possible Arrangement would be obb and then once again let's say somebody else comes along and says that you know what let's swap the Beast so this is what happens when you swap the Beast now you will notice you will notice that this is basically one unique Arrangement okay okay this is just one arrangement I mean it' be stupid to say okay by this is a different Arrangement this is a different arrangement I mean nobody can tell the difference and then this two also is one unique Arrangement okay so that's this is one way to arrange it this is the second way to arrange it and then this is the third and the final way that you can arrange the letters in the word b so the question is can we do 3 into 2 into 1 of course not can I do 3 into 2 into 1 = to 6 of course not so how do we do how do we deal with repetition repetition something is occurring twice so here's what we do what we do is 3 factorial that's the total divided by the number of letters that are being repeated which is 2 factorial and what's 3 factorial / 2 factorial that's equal to 3 okay so the answer is not 6 the answer is 3 factorial / 2 factorial and this 2 factorial is basically the number of in fact not number of it's the factorial of factorial of the number of same objects okay it's the factorial of the number of same objects yes that's correct okay so the number of distinct Arrangements of n object in a line of which P are identical to each other Q others are identical to each other all of a third type are identical to each other is so on and so forth okay so I'll write a note over here and the note is for Part B here it is the note is that when letters slash objects are repeated we divide by the factorial of the number of times they are repeated okay there you go why do we divide by the why do we divide the same letters you can see here because if we don't we end up counting the same Arrangement twice okay if we don't we end up counting the same Arrangement twice all right now part C trigonometry let's see let's see if you guys have learned the concept or not see if you guys can arrange trigonometry and tell me what the answer is the answer let me tell you is going to be a freakishly large value so please don't freak out it's prob it'll probably be correct if you've understood the concept you'll probably get it right I'll zoom out so you guys can see the working for part why don't you guys read the instructions I mean not the instruction sorry the title read the title you'll find the answer to your question you like the new can yeah thank you appreciate it the note is when letters SL objects are repeated we divide by the the factorial of the number of times they are repeated sir please regular streams for adms yeah yeah I will be streaming regularly like I said streaming binomial next probably after that I'll do trigonometry probably can I also use in class8 but why are you watching this I mean if you're in class eight Pokemon why are you watching ad maths in class 8 how much should I sleep for day I don't know at least 6 to 8 hours at least but it varies from person to person I would say 6 hours to 8 hours is enough all right anyone who's going to give me the answer to this or should I just solve it myself it's not the quantity of the sleep it's the quality of the sleep Sir could you please make a video on important topics all right the answer no I don't think uh anyone has given me the correct answer no so here's the answer so first of all let's count the total I'll explain nothing to worry about let's count the total total number of letters so we have 1 2 3 4 5 6 7 8 9 10 11 12 okay so that's going to be 12 factorial so 12 is the total now let's see what letters are being repeated so clearly there is T that is being repeated T occurs twice so we're going to divide it by 2 factorial so this 2 factorial is for the two T's that we have don't uh don't count all together the numbers are being repeated so we're going to do it separately for T separately for R separately for O then you have two RS okay so we're going to divide by two factorial so this is for the 2 RS and then we also have two o's okay so we also have to take that into consideration so this two factorial is for the two o's now if you work this out and given that you work it out correctly it's going to be 59 m875 200 okay so could you make a I did coordinate geometry of a circle I've completed it so you can check it out all right so now we're entering into some hostile territory which is is that what you said I don't think you said that anyway uh so we're entering some hostile territory where we going to have some restrictions and remember whenever there is a restriction we always make sure to settle or satisfy that restriction okay so if you have a restriction like for example a certain object should be in a certain place certain number of people should should be together certain number of people should not be together so you always first take that into account all right now the question says all right if you said the same good job it's difficult to keep track you know there are so many messages find the number of different ways in which all re find the number of ways in which all five letters in the word great can be arranged so how many letters do we have notice that we have 1 2 3 4 5 and all five of them are unique so the answer would simply be 5 factorial which is equal to what which is equal to 120 now part two so part two says how many of these arrangements are the letters A and E next to each other in how many of these arrangements are the letters A and E next to each other so there's a pretty standard way of solving questions like these okay there's a PR pretty standard way of solving questions like these and that is first the two letters that are supposed to be together let's put them in a box okay so here's a and here's e let's put them in a box so that we can make sure that they stay together now the question does not say that the letter e must be together in that order like you should have a first and then you should have e that means they can Shuffle amongst themselves it could be AE it could be e so that would be 2 factorial okay now make sure that you're paying very close attention now if you count the number of unique objects that we have to arrange so this will be treated as one object okay why because wherever they go they will go together so AE will be treated as one object so that means now we have to arrange how many different objects now we have to arrange one 2 3 4 we have to arrange four different objects how do you arrange four different objects in a straight line that's four factorial now let's see what that is equal to 2 factorial into 4 factorial that's equal to 40 8 okay so what's this 2 factorial for I'll write it down over here this 2 factorial is for A and E why because Inside the Box they could either be as AE or they could be as EA so we have to take that into consideration and what's this four factorial for this four factorial that you're looking at is basically the number of different objects that we now have to arrange the number of of different objects we have to arrange okay and remember AE will be counted as one okay now so why four factorial when three spaces are left I just explain because we have to consider 1 2 3 4 these are the four objects 1 2 3 4 that we now have to arrange what if they cannot move uh we'll talk about that we'll talk about that okay we're going to deal with different different instructions how do we use why do we use because they can be AE or they can be EA so how do you arrange two objects in a straight line two factorial now how many Arrangements of the word achieve are there if there are no restrictions on the orders the letters are to be in so first part fairly simple let's see how we can solve this so how many letters do we have in the word achieve Al together we have 1 2 3 4 5 6 7 so that's seven now watch out for any repetitions I can see e is being repeated so we have to divide by 2 factorial so 7 factorial divide by 2 factorial that's equal to 2520 butan whatever stream I'm doing I announce on my Instagram so you can check it out from there okay so and the rest of you I appreciate if you please stop spamming now part two part two says the first letter is in a now notice we have ourselves a restriction so what are we going to do we are going to first settle that restriction so whatever restriction you have you deal with that restriction first so we have a total of seven spaces make seven blanks representing the seven letters so the first letter must must be how many ways can we arrange this just one that's it okay so whether you write one or not makes no difference now what about the rest and remember a has to be in the first place it cannot be in any other place it has to be in the first place so that means we're only arranging the remaining six okay we're only arranging the remaining six and how can you arrange six letters six factorial but once again watch out for the repetition we have arranged a that means the two e are still in the pack are still somewhere over here so we have to divide by 2 factorial to deal with the repetition so 6 factorial divide by 2 factorial is equal to 360 way so there are 360 ways that you can arrange the letters in the word achieve given that is is in the first place now part three part three says the letters a and I are to be together so a and I must be together a i 1 2 3 4 5 6 7 okay so let's put a and I together in a box and because these are two letters and the question does not say that they must be in a specific order that means we have to consider how they can Shuffle amongst themselves that would be 2 factorial okay it's like two friends have decided that they're going to sit together so it doesn't matter doesn't like let's say there are three people who want to sit together in a plane so it doesn't matter all right so we have to consider all possible options so two factorial then this is going to be treated as one unit so that's 1 2 3 4 5 6 so 2 factorial into 6 factorial and remember remember there are are there are two letters that are being repeated so 2 factorial into 6 factorial IDE 2 factorial what's that equal to two or two cancel and 6 factorial wait a minute wait a minute I think say 2 factorial 2 factorial cancel 6 factorial is equals to 720 that is the answer okay a lot of people make this mistake that they don't like I can see some of you have also made that mistake they don't take into account that AI can also Shuffle amongst themselves okay and number two you also forget to take into account that this can be anywhere like you can have a letter first and then you can have ai together and then you can have the remaining letters okay so that's why we take this into account also that's why we take six factorial so we have to arrange six different objects now okay so just because two objects are together does not mean that they are going to be together in the first two places no they can be anywhere so that's why you have to take that into account also and that's why we're doing we're not doing five factorial we're doing six factorial all right so let's do part four how to solve without a calculator with a Common Sense 6 into 5 into 4 into 3 into 2 into 1 that's how you do it without a calculator now part four the letter C and H are to be a part now now the way that we solve letters c and HR to be a part is this and this only applies to two objects okay keep that in mind that two objects I'll write it down over here in fact two objects are not together the number of ways that we find that two objects are not together is equal to Total minus the number of w is two objects are together so it's kind of like probability probability of winning is equals to 1 minus the probability of losing okay probability of not raining equals to 1 minus the probability of raining okay that's how it works so total we already know that the total is 2520 now if you think about it here are our C and H the way that C and H will be together is going to be the same as the way a and I will be together so you don't really have to do it from scratch you could you realize it's the same thing so we don't have to do any separate working for this unless of course it was maybe with e because these are C and cnh don't occur more than once so that means the number of ways that cnh will be together will be the same as a Andi so we already have the answer to this the answer to this is 720 so 2520 is the total minus the number of ways that two objects are together and the answer is 1,800 so that will give us the number of ways that two objects are apart so why are we not multiplying two as AI can also be written as I I did multiply it by two I did multiply it by two but then they got cancelled okay in the working they got cancelled yes there should be a part there should be a part so I'll explain it again the number ofs two objects are not together is equal to Total minus the number of ways that two objects are together okay so either they're together or they're not together so instead of finding out the number of ways are not together we have already found out the number of ways they are together now you might argue that well we did it for a and I we didn't do it for C and H but they'll be the same because you know they're still two letters and none of these letters AI or CH are occurring more than once so what we're going to do is we are going to take the total and from it we're going to subtract the number of ways they're not together the number of ways they are together sorry and that way we'll get the number of ways they are not together it's like number can be even or odd if you know the total and from that you subtract all the odd numbers you have all the even numbers sir my ADM teacher said that repeating but I doubt your adma teacher said that I think you probably misheard him or misunderstood but see you don't have to rely on me or your teacher or anyone for that matter syllabus is not like a top secret FBI document you can just go on the website and just check it out for yourself okay or you can just open up a few P papers you'll realize how things work now this is something that I want you guys to make a note of you might come back to it I'll update the notes is because there a different this is how all three are together and of course we can consider the number of ways they can be shuffled okay not all three are together means that two can still be together okay not all three are together means that two can still be together no two are together means that all three must be separated like you can't even have two together okay why because sometimes people mix up this with this okay people usually mix these two up so it's important to understand the difference all three are together is quite straightforward but not all three are together means that less than three two can still be together but no two are together means that however many they are no two of them can be together so that means if there are 10 all 10 must be separated okay that's what it means anyway let's do some more questions and uh I don't think we'll be able to complete it by today but let's see let's focus on quality over quantity three girls and two boys are to be seated in a row so right now we're doing questions from the book okay we're not doing possible questions we'll start possible questions soon find the number of different ways this can be done if the girls and boys sit alternately alternately that you have a boy and then you have a girl and then you have a boy and then you have a girl like that okay so we have a total of five people 1 2 3 4 five now notice the question doesn't specify whether you should have a boy first or you should have a girl first so is this possible that we have a boy a girl and then a boy and then a girl and then a boy what do you guys think you think I can do this yes or no read the question carefully and then tell me why not reason why this is not possible no this is not possible because if you look at the number of boys there are only two boys and as for this arrangement in order to do this arrangement we need how many boys we need three boys well so this is not possible even though they're sitting alternately but this is not possible to begin with because we don't have that many number of boys so alternatively what we're going to do is this we're going to start the seating arrangement by having a girl sit first and then a boy so girl boy girl boy and then girl okay so let's focus on the number of girls that we have so notice that we have three girls so that means we have three options here okay so we have three options here we have two options here we have one option here and then if you talk about the boys so we have two options here and then we have two options here and if I want to find out the seating arrangement Al together we'll just multiply and let's see what do we have 3 * 2 is 6 6 * 2 is 12 12 * 2 is no wait 3 * 2 is 6 6 * 2 is one now it's correct so 3 * 2 is 6 6 * 2 is 12 so that means all together there are 12 ways that you can arrange three girls and two boys given that this sit alternatively no it's not 24 sorry my bad I made an error because there are three girls girls then two then one then you have two boys and then one boy there's no point of me repeating it unless it's difficult so I'll in that case I will summarize it multiple times now Part B A girl sits at the end of each row now where is the Restriction the Restriction is at the end of each row so we will first make sure 1 2 3 4 5 yeah we'll first make sure that there's a girl sitting here and there's a girl sitting over here now how many options do we have for the first place we have three options okay because there's got to be a girl right and since there's got to be a girl at the end of the row that means we have two options here now in between Anything Can Happen nobody really cares so that means we started with five people now from the five two of them are gone okay remember there were five people Al together two of them are gone now we're left with three so that means in between we can have three factorial why three factorial because there are three people that we're left with and we have to arrange all three of them so that would be simply 3 factorial so all together it's going to be 3 into 3 factorial into 2 so that's 6 6 * 2 that's equal to 36 so there are 36 ways okay now part C part C says the girls sit together and the boys sit together all right so if the girls sit together let's start by making five blanks representing the number of people that we have 1 2 3 four five so the girls sit together now this means that the number of girls can Shuffle amongst themselves however way they like so since there are three girls that would be three factorial same goes for the number of boys I mean as long as they're together it doesn't matter that means that would be 2 factorial okay so the answer will be 3 factorial into 2 factorial correct 3 factorial into 2 factorial you guys think will be the answer No 3 factorial into 2 factorial is not the answer nope 3 fact into 2 factorial is not the answer why it's going to be 3 factorial into 2 factorial into 2 factorial why into 2 factorial let's understand so the 3 factorial is basically for the number of girls this 2 factorial is basically for the number of boys okay so this these these are the girls these are the boys now the question does not say that you have you should have girls first and then boys it what if it's boys first and then girls does the question doesn't specify so this two factorial is basically representing that you can have all the girls first and then boy or it's also possible that you have all the boys first and then girls okay so altogether the answer will be equal to 24 so there are 24 ways that you can have all three girls sit together and all two BO sit together okay so we now have ourselves a pause paper question finally find the number of different Arrangements of the letters of the word Mexico so the first part you will notice is usually going to be without any restriction now there are six letters in the word Mexico none of them is being repeated so 6 factorial is equal to what 360 no it's not 3 yeah it is 720 sorry an advice would be for you guys to memorize all factorials ideally from 1 to 7 to save time you can always work it out there and then but make sure to memorize them to save time now part two part two is find the number of these Arrangements which begin with the letter M so M condition M fix here are three blanks we need six so let's be let's work smart duplicate up in the first place there's got to be M that means we can Shuffle the remaining five places so that would be 5 factorial and that's it the answer is just 5 factorial which is equal to what which is equal to 120 ways 720 ways 120 ways now here's part three part three says which have letter X at one end and letter C at the other end so letter X at one end and letter C at the other end so here's X and C that means in between Anything Can Happen doesn't really matter so that would be 4 factorial is that correct that would be the answer four factorial letter X C what you guys think 48 smarty Hass that's correct good job so this is incorrect because the question does not specify that you can have X and you should have X in the first place and C in the last place it should be 4 factorial into 2 factorial which is equal to what which is equal to 48 two factorial you might argue it's the same but uh yeah yeah you're you're right basically chemtics stream let's focus on what we're doing I I do have a lot of videos already on kinematics if you're if it's urgent you can check them out but I'll make a note of it okay I'll tell my team to make a note of that now a fashion magazine runs a competition in which eight photographs of dresses are shown lettered ABCDE e f g competitors are asked to submit an arrangement of five letters showing their choices of dresses in descending order of the Merit the winner expect at random from those competitors whose Arrangements of letters agrees with that chosen by a panel of experts okay uh these are pass papers Abdullah what we're solving are pass paper questions okay calculate the number of possible Arrangements of five letters chosen from the eight okay now don't get confused by the word chosen we're still arranging them okay so these eight have already been chosen now we have have to arrange five of these eight so how can you arrange five objects or five letters from eight that would be simply 8 P5 which is equal to 6720 that's correct 6720 good job Alina watching from Zimbabwe all right that's great part two part two says in which a is placed first so now we have to make sure that a is placed first so let's see how we can do that these are five objects okay keep that in mind these are five objects so we're going to place a first okay now what about the remaining four in the remaining four we're not going to do four factorial okay fix in the remaining four what we're going to do is remember we started off with eight photographs okay we started off with eight photographs now of the eight one doesn't matter whether it's actually no it it is a a has already been arranged okay a has already found its place now we have to arrange remaining four from the remaining seven okay we have to arrange remaining four from the remaining seven so how can you arrange seven objects how can you arrange four from Seven objects that would be 7 P4 okay not 8 P4 7 P4 so 1 into 7 P4 which is equal to 840 they if you already arranged one object that means from the total that's no more part of the total right all right now part three which contain a okay now which contain a tell me an easier way tell me an easier way that we can have sir I'm curious what grades did you get in oh okay we'll talk about that later so what we can do 4200 that is correct good job that is correct so here's what we can do will this be yeah yeah this will be available for practice okay this will be available for practice now there are two ways that you can solve this okay there are two ways that you can solve this yes one is 7 P4 into 5 okay good job the other is this what would be the opposite what would be the opposite it would be no a okay total minus no a will give us the number of Arrangements which contain a okay so total I think this is the one that's going to make the most sense to you guys so bear with me if I have the total number of a which contain a and from that I get rid of all the possible Arrangements that do not contain a what will happen what will happen is we will have the number of Arrangements that contain a okay but if you didn't attend the first 20 minutes just go back and revise I mean you can always remind it's it's a live stream you feel free to reind okay so total we already know what the total is that's 8 P5 okay and from this we can subtract the number of an Arrangements that do not contain a now how will I figure that out how will I figure out the number of Arrangements that do not contain a that would be if we make sure that we don't select a okay if we make sure that we don't choose a or we make sure that we don't arrange a so that means from the eight we're not going to arrange all eight of them we're only going to arrange seven of them so that a arrange and we're going to arrange five of them okay so if let's say I ask you to find out the number of possible Arrangements of five letters which do not contain a so you're just going to get rid of a to begin with and you're going to arrange the remaining seven you're going to arrange five from the remaining seven okay so 8 P5 - 7 P5 let's see what do we get so that's 6720 - 7 P5 let's see what that is 7 P5 that's 2520 the answer should be 4200 6720 - 2520 yep 4200 is the correct answer so why why can't we directly find it see the way that we do it directly is a little bit tricky which is why I didn't explain that I will tell you how to do it so how many ways can you have a in the first place that's 840 but a will not be in the first place a could be in the first second 3D fourth fifth so an alternative way now if you get it that's great if you don't I would suggest you stick to the approach above alter alternatively here's what you're going to do 840 is the number of ways that you can arrange a in the first place but a will not be in the first place a could be anywhere in the first place so 840 into 5 that's equal to 4200 but I'll put a star over here if you get it that's great if you don't nothing to worry about uh why 7 P5 because how do you make sure that you don't end up arranging a you only arrange from the seven which do not contain a it's like you're selecting a team and you want to make sure that one person does not get selected so you're just going to remove that person from the pool okay yeah APG part nothing to worry about okay now we're entering some even more hostile territory and this is when we apply permutations to numbers this is where things can get tricky but as always if you follow instructions and if you do things step by step this can also be a piece of cake for you okay find the number of different four-digit numbers that can be formed using yes that's correct 814 to 5 to place it in other places find the number of different four-digit numbers that can be formed using the digits 3 5 and seven without repetition so notice that we have four numbers that we're supposed to arrange and we can do it in whatever order that I want so 4 factorial that's equal to what 24 no repetition to worry about and this is without repetition without repetition means that if you've used the digit once you cannot use it again okay normally when we're setting up passwords that's not how it work that's not how they work we can you know password 1 one11 0 but not in this case now Part B B part one says how many of these four-digit numbers are even now remember the rule is that whenever there's a restriction you settle that restriction first so in order to make a number even I'll make sure that the last digit is eight okay this doesn't mean that I have eight options for the last digit that means the last digit is eight which means I just have one option three numbers could be or doesn't matter the number will be even as long as the last digit is even doesn't matter whether it's 200 200 2,348 or 3,428 or 5,728 doesn't matter the number will be even so in between anything can happen that means there are three numbers that can Shuffle amongst themselves nothing doesn't matter so the answer is 3 factorial into 1 which is equal to 6 now part two greater than 8,000 now if you want to make sure that the number is greater greater than 8,000 you will make sure that the first digit is 8 okay so you have just one option for the first digit and the the three that follow doesn't matter what happens P digit eight here the number and given that there are no zeros the number will inevitably be greater than 8,000 so 1 into 3 factorial that's equal to 6 okay yeah same answer now you might think you know this piece of cake but you will be wrong why because of some questions that I'm about to solve so here's a question that I would like to solve with you guys yeah this is easy AB just take it easy you will find out when things get difficult so this was one example of permutation being applied to numbers here's another example okay and this is a PO prer question so it says a six-digit number is to be formed using the digits 1 3 5 6 8 and 9 each of these digits may be used only once in any six-digit number find how many different six-digit numbers can be formed if there are no restrictions so no restrictions that means there's nothing to worry about there are six numbers we can arrange do we have to arrange all six of them yes so the answer would simply be 6 factorial which is equal to 720 okay now let's talk about Part B the number formed is even 1 2 3 4 5 6 so if the number formed is even that means the last digit could either be six or eight so that means we have how many options we have two options for the last digit okay not two factorial two factorial that's not the case we have two options and but we have only room for one so it's just going to be two or if you want you can do 2 P1 but I wouldn't suggest that just keep it at two now here for the remaining five places remember we started off with we started off with how many numbers we started off with 1 2 3 4 5 6 we started off with six numbers now of the six either six or eight will be gone okay so if 6 or eight is gone that means we are left with five and we have to arrange all five of them so arranging all five numbers can be done by five factorial so the answer is going to be five factorial into 2 which is equal to 240 so 5 factorial into two which is equal to 240 okay not 2 factorial please note that we will not have 2 factorial over here so that you guys know that 2 factorial instead you will have two why because either six eight you have two options just like if you go back to which question the girls and boys question like here we didn't do three factorial we had three options so that's why we did three and there was only room for one okay now so far so good I hope now these are the type of questions that I love the one that I'm about to solve the number formed is even and greater than 300,000 so let's make six blanks 1 2 3 4 5 6 part three is 120 no part three is not 120 I would suggest that you guys pay close attention when I'm solving part three I'll do another example similar to this but make sure you're paying attention so let's try and make sure that the number is even that means we could have six or eight in the last place and then let's try and make sure that the number is greater than 300,000 that means it would it could be 3 5 6 8 or 9 in the first place okay so I repeat the number could be 6 or eight in the last place or in the first place we could have 3 5 6 8 and N now there is a problem over here who can identify what that problem is there is a problem what is that problem and identify that problem nope it's not 240 I'm not interested in the final answer okay I'm not interested in the final answer yes same digits on both ends now I like to call these situations conflict situation okay so this is a situation where you have a conflict you have a conflict yes there is this is a conflict situation the conflict situation is between six and 8 okay that you have a six here and you have a six here or you have a eight here or and you have an eight here that doesn't make sense so at one at at a time there can only be one digit in one place so how do you deal with these questions you deal with these questions by separating them okay where is the conflict the conflict is between 6 and 8 yes so what we're going to do is we're going to fix six and then we're going to fix eight so here's one way that we will fix six as the last digit so let's fix six as the last digit let's see what the situation then looks like 1 2 3 okay if I do if I fix six in the last place that means I have one option okay now remember you still have to make sure that the number is greater than 300,000 so now you can have either three or five or 8 or 9 now there is no conflict okay everyone every space has its own set set of unique numbers so there's no conflict at all and in the first place that means we have four options not four factorial we have four options now in between doesn't [Music] matter we don't really care so notice that we had six numbers of the six one will go in the last place one will go in the first place that means in between we have four spaces and we are left with four numbers as well so that would be four factorial okay I repeat we start off with six numbers of the six one two of them are gone so that means now we're down to four and we have room for four also so that would be 4 factorial so what's this equal to 4 factorial I think that's 24 but let's work it out 4 factorial into 4 that's 9 6 okay this is not the final answer now we have fixed six as the last digit let's fix yeah you can write 4 P1 you can write 4 P1 and you can write 1 P1 that's perfectly all right let's fix 8 as the last digit so if I fix eight as the last digit that means now we can have three 5 6 or 9 as the first digit no the answer is not 240 who said the answer is 240 it's not 240 so that means we have four options for the first place one option for the last place in between we are down to four numbers and we can arrange all four of them however way they want so this would also be equal to 96 and now the total the total is equal to 192 and this is the final answer while this is fresh in our memory I would appreciate that we immediately solve another question okay where is the other question just a second easy J I'm glad you find it easy I'm extremely happy that you find it easy this is like probability um well if you think it's like probability then fine yes if you have taken six in the last place you can't take it in the first place you can't have one six in multiple places uh so I have done a detailed stream on permutations and combinations for AI so I would suggest that you check it out please okay now just a minute let's solve this question yeah let's see what this says so this question says six-digit numbers are now find how many different four-digit numbers can be made using the digits 2 3 5 7 and 7 8 and N if each digit may be used only once in a number so remember here we have let's count the number of digits that we have 1 2 3 4 5 6 so we have six different digits and we only have to make a four-digit number that means we only have to arrange four of them so that would be 6 B4 that's equal to 360 good job you guys are getting the hang of it that's nice how many of the numbers found in part one are divisible by five now how do I make sure that the number is divisible by five how do I make sure that the number is divisible by five what do I do should I place two as the last digit one as the last digit five as the last digit Z or five you're right Z or five but okay so that means we can only have five as the last digit okay so let's have let's fix five as the last digit that that means in between anything can happen you know nobody really cares so how many numbers do we have we have a total of 1 2 3 4 5 six 6 now we're down to five and we have to arrange three of the five numbers so that would be 5 P3 is 5 P3 5 P3 into it's going to be 5 P3 into 1 which is equal to 60 that's correct 60 numbers all right all right part three my favorite how many of the numbers found in part one are odd and greater than 7,000 so here's what you have to do you have to attend or you have to attempt this question without knowing that this is a I mean right now you guys probably know this is a conflict situation but when you're doing a question like this in the exam just try and do it the normal way if you see one number which can potentially be in two places that means it's a conflict situation so we have to make sure that the number is or that means we can have 2 3 5 7 or 9 okay and we also have to make sure that the number is greater than 7,000 which means we could have 7 8 or 9 as our first digit that means now there's a conflict now where do we have fewer numbers we have fewer numbers in the first place so that means we're going to fix seven in the first place we're going to fix then eight in the first place and then nine you could do the same you could do do the same by fixing 2 3 5 7 9 in the last place but here you'll have to do it five times because there are five options here there are three options not two oh yeah of course not two number are two thank you for pointing that out 3 5 7 and N 35 7 9 nonetheless uh thank you for pointing that out okay so let's fix seven so this is because of Seven it's a conflict situation so this is a conflict situation okay so let's fix seven in the first place not just because of Seven actually I just realized because of seven and 9 why do we have to go for eight because the number should be greater than 7,000 Okay so so the number should be odd and at the same time greater than 7,000 so if I have 7 that means I can have three 5 or 9 so that's one option here and three options here now remember we had six digits to work with one and two two of them are gone and now we're down to four and we have room for how many we have room for two so 4 P2 okay so let's work this out see what do we get 4 P2 into 3 that's that's equal to 36 okay now let's fix 8 as the first digit with practice realize numbers working same so you won't have to do it every single time so if I fix eight as the first digit so that's eight now we can have three 5 7 or even nine yeah we could have 3 5 7 or even nine as our last digit that means we have one option here we have four options here so here we have once again we started off with six numbers of the six two are gone now we're down to four and we still have room for two so that would be 4 P2 so 4 P2 into 4 that's equal to 48 now let's fix N9 as the first digit 20 how can 20 be the final answer we already exceeded 20 how can 20 be the final answer so now let's fix N9 as our first digit that means now we can either have three five or S as our last digit so we have one option here three options here sorry not three factorial and in between it's going to be 4 P2 same as above so this will also be 36 now let's add them up see what do we get so it's 36 + 36 + 48 that's equal to 120 so there there are 120 numbers that's equal to it will be nine8 already so yes 120 is the correct answer yes last one will also be 36 and therefore the answer is 120 okay so I hope you guys are getting the hang of it now let me skip a few questions and save some time skip if we can okay let's do by the way there are some worksheets that you guys will find yeah you can do it by fixing the L number as well the reason why I didn't opt for that is because we have fewer options in the first place and more options in the last place yeah you could do that okay so so let's solve a few b Paper questions uh those kind of questions where you do permutations without a calculator I will solve separately okay I'll make a video on it and share it with you guys separately now here's a question let's solve this and then there are some other questions also that we are are going to solve okay and I would suggest you guys to do the worksheet not right now later on you'll find a plenty of worksheets make sure that you guys do those worksheets as practice so please let us go at 10 well you guys are free to go now let's do question number8 this this is a question that's from the worksheet okay so this question says a five character code is to be formed from the 13 characters below shown below each character may be used only once in a code find the number of different codes in which no two letters follow each other and no two numbers follow each other so that means if you have a letter you got to have a number first and then you got to have a letter first okay so we have to make a five character code let's see one 2 3 4 5 let's say we start with a number okay how many numbers do we have we have seven numbers 25 no it's definitely not 25 okay is this per this is permutation and is this permutation and combination those of you are asking this is definitely permutation because we're arranging them okay now 25 24 is not the answer please pay attention so we can have a number here then we can have a letter then we can have a number and then a letter and then a number again okay so let's see what happens how many numbers do we have we have seven numbers then let's deal with the numbers first okay then we have if we use one number that means we can only use six the remaining six and then we can only use the remaining five and now let's talk about the letters so we have five letters so that means we have five options for this place and then we have four options for this place okay so let's work this out we have 7 into 5 into 6 into 4 into 5 so what do we get we get 4200 okay so the answer is in this case we get 4,200 all right I'll repeat so we have a number in the first place and we have a total of seven options and then we have six options and then we have have five options okay and then as far as okay I just realized that we don't have we don't have five letters we actually have six letters sorry we just we have six letters nobody pointed that out interesting so we have six letters here and then we have five options here so this is not going to be 4200 let's see what this is going to be this is yeah thank you for pointing that out 7 into 6 into 6 into 5 into 5 that's equal to 6,300 now notice that this time we started with a number what if we start with a letter the question doesn't specify that you have to start with a number we could also start with a letter so let's try that let's try that that we start with a letter so letter number letter number letter okay remember it's a five character code no no it's not going to be 6300 * 2 I'm afraid it's not going to be 6300 * 2 because we don't have an equal number of letters or an equal number of numbers that's not the case we have six letters and seven numbers so you can't do simply 6300 * 2 and you will see why you can't do that so letter number letter number letter now how many letters do we have how many options do we have we have 1 2 3 4 5 6 so that's six and then five and then four and then when we're talking about numbers that's s and six okay so let's work this out see what we get 6 into 7 into 5 into 6 into 4 that's equal to 50 40 so all together all together it would be 6300 + 504 which is equal to 11,340 there we go okay now let's do another pause paper question a computer password which must contain six characters is to be chosen from the following 10 characters each character may be used only once in any password find the number of possible passwords that may be chosen if each password must contain at least one symbol okay at least one symbol what would be let's see how many ways it can have at least one symbol it can have one symbol two symbol three symbol so that means that's a lot of working okay so what would be the opposite of at least one symbol we will do combinations but uh let's let's do a few I think this is the last permutation question after this we're starting combination okay so tell me how many uh what would be the opposite of at least one number so if let's say the condition to get into a certain college is that you should get at least one a then what do you have to do in order to make sure that you don't get into that college no symbol that's correct so if the condition to get into a certain college is that you should get at least one a the only way that you won't get into the college is by if you don't get an A at all okay so the opposite of it at least one would be none so keep that in mind so we're going to do this question smartly so total minus no symbol will be equal to at least one symbol all right total minus no symbol would be equal to at least one symbol now notice that there are 10 characters and the password must contain six so that means total without any restriction would straight up be 10 P6 okay minus no symbol now how do you make sure that you form a password which has no symbol how do you make sure that you form a password which has no symbol so if I'm making a password which has no symbol that means how many options do I have to choose from I have I'm just going to cross out the symbols so I have 1 2 3 4 5 6 7 I have seven options to choose from and the password that I'm making must have six characters so that's 7 P6 let's see what this is equal to so 10 P6 - 7 P6 and that's correct the answer that we get is 146,000 160 there you go makees sense I hope okay up finally start combination let's put this in trash now let's start with combination I didn't get it uh okay so at least one symbol and we said that you know what let's find out the opposite of what the question is saying that is let's find out the number of passwords that we can make that have no symbol so we said 10 P6 that's the total from that if you want to make a password that has no symbol that means don't use a symbol if you want to make a password that has no symbol then don't use a symbol when making a password use the remaining seven characters so from the seven characters we're going to use six so that's 7 P6 so 10 P6 - 7 P6 = to 146 1 160 yes you can watch for s one but for S1 I would suggest that uh I have a dedicated playlist of permutations and combinations just may I have done permutations I've done one video of combinations but you can check it out you can still check it out all right now let's talk about combinations now what are combinations combinations are basically selection I mean you use combination when there's selection or you're making a team or you're simply choosing in which case order is not important something that we discussed earlier so if you have a and b that's the same as b and a why because it's the same two people therefore it's the same team and if you're in a situation where AB is the same as ba that means it's a combination question okay so let's do a few examples where there are no restrictions now you can see in this question it says how many different ways are there of selecting so the keyword is selecting I'm select that means we have to use combination now what are we selecting we're selecting three photographs from 10 photographs pretty easy three photographs from 10 photographs that's going to be 10 C3 okay so in combinations what we use is we use C this is the formula but don't worry about this there's a paper one part of it which I will make a separate video on you guys don't have to worry about it right now so 10 C3 is equal to what is equal to 120 then 5 books from seven books that would be 7 C5 which is equal to what which is equal to 21 then a team of 11 footballers from 14 footballers that would be equal to 14 C1 let's see what that is equal to use your calculator 14 C11 that's equal to 364 I will teach you guys how to do NPR and NCR without a calculator but not right now okay now how many different combinations of the three letter words can be chosen from 5 C uh from the pqr St so we have five letters and we have to select three so that's going to be 5 C3 and what's that equal to that's equal to 10 pretty simple okay now we're going to gradually increase the difficulty level okay an Athletics Club has 10 long distance Runners eight sprinters and five jumpers a team of three longdistance runners so we have to select three from 10 okay five sprinters from eight and five jumper or two jumpers from five are to be selected so how can we do that let's start with 10 longdistance Runners from which we're supposed to choose three eight sprinters from which we're supposed to choose five five jumpers from which we're supposed to choose two okay so let's work this out 10 C3 into 8 C 5 into 5 C2 that's equal to 67,2 a what's your question sir in igcc I'm in 10th and chosen ad maths should I choose math a level yeah you should definitely choose math a level I would strongly recommend it so here you are selecting basically from three categories now I don't recommend further math uh if you haven't been challenged by math at all then I would recommend otherwise I don't so from the 10 long distance Runners you're supposed to choose three how can you do that that's 10 C3 and remember this entire event will be completed when you select longdistance Runners sprinters and jumpers okay so which is why we are multiplying so we're within the same event that's why we're multiplying okay so I've I've answered your question it's the same event it's like the first is blue and the second is red if you remember math probability so we do blue probability of blue times the probability of red so event if there were multiple ways of let's say fulfilling the same condition or selecting a team under the same restriction then we would add okay we're going to talk about that so 10 C3 this is how you select three longdistance runners from 10 8 C5 how you select five sprinters from eight 5 C2 is how you select two jumpers from five and we multiply because all of them together make a club and that's how you get 67,2 sir can you do as yeah yeah I will do as as well don't worry okay inshah I'll be streaming regularly from now on four of the letters in the word paintbrush are selected at random find the number of different combinations if there is no restriction on the letter selected so here you can see that there is no restriction in the first part so how many letters do we have in total let's see 1 2 3 4 5 5 6 7 8 9 10 so that's 10 from the 10 we have to select four so that's 10 C4 what's that equal to 210 okay A Part B Part B says the letter T must be selected so here's an important note if T is already selected then we are then we are selecting from already that means remaining letters select that means we're now selecting from T is gone okay so we are now selecting from painbrush okay so we are now selecting from paintbrush now how do we do that well now we only have nine to select okay nine to select from and from the nine we only have to select three okay now because T has already been selected so if you want to understand the logic behind it so 1 C1 this way we will make sure that t must be selected because there's one t and we have to select it so that's 1 C1 okay and this is how we're selecting the remaining okay that's not how you spell remaining remaining three from the remaining nine okay so let's see what do we get 1 C1 into 9 C3 that's 84 so you can list down the keywords they select choose Team all these word all these words are cues that you have to use combination sir the order doesn't matter it's just about picking things that's correct when you are just selecting the order doesn't matter okay now let's do these questions and then I think we will have a separate stream for Boss paper question okay I don't know about you guys but many bus okay now question two a committee of five people is to be chosen from Six Women and send seven men find the number of different committees that can be chosen if there are no restrictions okay no restrictions let's see so we have a total of 6 + 7 which is equal to 13 and from the 13 we have to select five so that's 13 C5 what's that equal to that's equal to okay I'm not that smart 13 C5 that's 1287 1,287 okay now let's do Part B if there are more men than women so we have to select a team of five and we have to make sure that there are more men than women so remember that there are seven men to choose from and there are six women to choose from okay so how can we make a team of five which has more men than women we can have three men and two women how do you select three men from seven and how do you select two women from six we know how to do that right this would be 7 C3 into 6 C2 okay so 7 C3 into 6 C2 we can work that out later if anyone can work this out that would be great we'll save some time now is that the only way that we can have a team which has more men than women no that's not true we can have four men and one women that would be 7 C4 into 6 C1 and then 525 all right thank you another way is five and zero okay you still have more men than women right you might argue well they're all men but literally speaking it's still more men than women there are five men zero women so there are more men so this can be done 7 C5 into 60 Z okay so 210 here to 21 so this is going to be let's see 5 + 1 let's just use a calculator 756 yeah that's correct 756 okay so an important note there's always a note and here's the note the note is that since there are multiple ways of meeting the same requirement so there's a requirement and we can meet it multiple ways then that therefore we add okay which school am I I'm not teaching in any school I am teaching online entirly okay let's do one more question and pause paper questions inshallah we'll solve in the next stream which will be which will be I'm thinking maybe Friday let's do it on Friday because Thursday I'm already streaming so it can't be tomorrow tomorrow I'm busy but uh let's do it on Friday when do we multiply like for example we when we're forming the same committee like these two together will make one committee so that's why we will multiply okay now it's the same it's the same requirement which can be fulfilled number of ways so I mean there are multiple ways of fulfilling the same requirement so that's why we find out the number of different committees we can make which fulfill the requirement and that then we add okay question number three example number three sorry a team of five members is to be chosen from five men and three women find the number of different teams that can be selected if there are no restrictions so we have to select five from eight so that would be 8 C5 yeah most likely on Friday okay most likely on Friday so 8 C5 that's equal to 56 that consist of three no no okay I won't delay it that much Three Men and two women so how do you select three men from five that would be 5c3 how do you select two women from three that would be 3c2 let's see so 5c3 into 3 C2 that's equal to 30 that consist of no more than one women all right this person seems to have a bias against women sorry about that but uh that consist of no more than one women let's see how many ways we can make a team of five that consist of no more more than one women so we have remember we have five men to choose from okay and we have three women to choose from now how can I make a committee of five or a team of five which has no more than one women that means we can have five men and zero women okay so how do you select five men and zero women that would be 5 C5 and 3 c0 which is equal to one or you could have four men and one women okay no more than one women means that you could have zero women or you could have one women so this would be 5 C4 into 3 C1 5 C4 is 5 3 C1 is 3 so that's 15 15 + 1 is equal to 16 so you can make 16 different teams which have no more than one women 10:30 what do you guys suggest I think let's stop let's that way we'll be able to do more passer questions in the next class inshallah okay so in the next class there are some uh interesting questions that we're going to solve you know especially those questions where you have two twins that must not be separated a husband and wife that must not be separated a husband and wife that must be together or one twin gets selected stuff like that so we're going to do that inshallah in the next class okay some fun questions which we'll solve now before you guys go I would appreciate if once again you haven't liked the stream do that and also subscribe to the channel if you haven't done that sir if I solve the worksheet from those are possible questions M those are possible questions where we have I my team and I have basically picked specific Concepts okay so those are passp paper questions you will see the reference also if you can do those you can you are doing passp paper questions yes I do plan on making that I'm working on it it's uh I mean it's one of those videos where I have to I want to make sure that I get it right so I have to make it responsibly so you'll find it inshallah maybe in maybe this week or the week after you can download the worksheets from the link in description please download the worksheets from the link in description anyway sir can students join your team to in become interns well when there is a vacancy you will find on my Instagram okay if there's a vacancy you'll find it on my Instagram but uh I'm sorry there's no such thing as an internship okay because it's all remote work so there's no such thing but if you're if you're interested you know you can follow follow mat lead byad on Instagram if there's a way can see you can apply anyway so I'll stop here fellas see you guys on Thursday Thursday I'm streaming binomial Friday most likely I will do a part two of this okay so take care once again Allah