Okay, let's try the question of IIT-JEE. The question is, if RST are prime numbers and P and Q are positive integers such that LCM of P and Q is R square T power 4 S square, then we have to find number of ordered pairs P and Q. Okay, P and Q are two numbers whose LCM is R square T power 4 S square. And the best thing is that RST is a prime number. So, we can take the distribution of RST separately. Now, let's first look at R square. The LCM of these two numbers should be R square. So, how will the distribution of R factor be in P and Q? Now consider that suppose the LCM of two numbers is 2 square. So how can the distribution of these two factors 2 and 2 be here? Here we have R instead of 2. How can the distribution of R be here? It is possible that R is not there even once in P. But at that time what should happen in Q? R square should happen. Because if LCM is R square, then one of these two numbers should have R square, otherwise LCM will not be R square. What are the other possibilities? Here R power 1, so what will happen here? R square. Here R square, so what will happen here? R square. Then also what will be its LCM? R square. What are the other possibilities? Here we will... R2 here is R1 and R2 here is R0 So here we have 5 possibilities for R We can write it as 2 x 3-1 I have written 2 x 3 because You can double the number of cases When you double the number of cases, how many times will this case come? It will come twice But we need this case only once So I have minused it Minused case and for r square we consider three possibilities of r power r power 0, r power 1, r power 2 similarly here we have t power 4 the possibilities of these powers will be five possibilities 0, 1, 2, 3 and 4 so the distribution of t power 4 will be 5 multiplied by 2 minus 1 and this will be 9. So, in this way, there will be so many possibilities for S square also of distribution of S factor that will be 2 multiplied by 3 minus 1 that is 5. Now, this is the S factor. are key possibilities are you T key possibilities are you a ski possibilities are so total number of ways total number of Joe options say kya ho jayenge 5 x 9 x 5 and that is 2 to 5 to hamara audit best kitne bhenge to 25 dance samosa of equation me hamle are ST ka Joe factors say Oscar distribution a la galaxy Q Kia Why is it possible? Because what type of numbers are R and S? Prime numbers. The equation is on LCM. So we should know that how the distribution of factors in LCM is done. If R square is coming in LCM of two numbers, it means that in any one number, R power should be 2. In LCM, T power should be 4. So in any one number, T power should be 4. Like that. So, if we look at this question to do it, then no theorem is coming to work of permutation combination. This whole question is running on logic. How much you can think, according to that you proceed here. Let's move on to the next question. Okay, now let's see selection from. identical objects. Suppose there are n identical objects, how many ways you can select one object? If all n objects are the same, how many ways you can select one object? One. Here I write number of objects and number of ways. Suppose I want to select two objects. If n objects are identical, then what is the way to select two objects? Close your eyes and select any two objects. So what will be the number of ways? 1. If there is 3, then also 1 will come. If there is 4, then also 1 will come. If there is n, then also what will come? 1 will come. If you want to select 0 objects, then also what will be the number of ways? 1 will come. So suppose you are told, in how many ways you can select at least one object from n identical objects. So, number of ways, what will happen? At least one means, either you select one in one way or you select two again in one way or you select three or you select n objects. So, how many times will it be 1, 1, 1? It will be n times. Suppose, you are told, how you select 0? or more objects. Now, we have to include zero selection as well. So, this 1 plus 1 plus 1, how many times will it come? It will come n plus 1 times. So, the answer will be n plus 1, which is included in the case of zero selection. Now, from this concept, there was a question in Okay, the question is, there are 10 identical white balls, 9 identical green balls, and 7 identical... black balls and we have to select one or more balls we have to find number of ways that means we are selecting at least one ball Now suppose we select only one ball, then from where can that ball be? What color can it be? It can be white or green or black. If it is white, then nothing comes from green and black. If it is green, then nothing comes from white and black. Like that. So what does it mean to say that this selection is also possible in such a way that some color balls are not selected So what we will do in this is that we will count the total number of possibilities separately for each color ball What are the possibilities for 10 white balls It is possible that not even one white ball comes or one or two or all white balls come So how many times will 1 plus 1 plus 1 come? 11 times 9 green balls so how many cases will be made here? 10 cases 0 green, 1 green, 2 green, 9 green and for 7 black balls, 8 possibilities will be made now in your answer, one case will be such that nothing will be selected it will be one case only so we will minus it because we have to select something so when we will simplify it, its value will be this is 880 minus 1 this is 879. So, look at the selection of identical objects, good equations are made. And the selection of identical objects, the most important concept is the concept of divisor. You know how to find divisor of any number. So let's see some questions in which we have to find the divisors. For the number of divisors, we will consider a number 24. We will express it in the form of prime factorization. Like this. Now we have 4 integers products here. 3, 2, 2 and 2. And we have to make a divisor. So what we will do for the divisor is to select something from these 4 objects. Suppose I have selected 2, then the divisor is 2. I have selected 3 and 2, then the divisor is 6. If I am not selecting anything, then what will be the divisor? If I am selecting all, then the divisor is 24 itself. The number itself will be. So, here there is selection, but some of the objects are identical to me. So, I will make two groups of 3 and 2. Now, from here my selection will be from 2 ways, it may be that 3 does not come once or 3 comes once. And what will be the number of ways of selection from here? How you select 0 or more objects from 3 identical objects? Number of ways will be 4, it may be 2. If 2 is not there, then no matter how many times we select 2, the number of ways will be 1. So the total number of ways is 1 plus 1 plus 1 plus 1, 4. So the number of devices will be 8. We will multiply these two because the logical operator between them is the end. So, what is the formula? 1 is 1 plus 1, 3 is 3 plus 1. And anyway, you check what are the 24 devices? 1, 2, 3, 4, 6, 8, 12, 24. You can check that these are 8 devices. Now let's take a general number. Suppose 2 power p, 3 power q, 5 power r, 7 power s. These are all prime factors. So the number of devices of this will be p plus 1 multiplied by q plus 1 multiplied by r plus 1 multiplied by s plus 1. Suppose we need even divisor, for that we need two factor at least one time. So at that time, two possibilities will be P, one will be less because zero selection is not allowed here. And the other options will remain the same. If I want more divisors, then it means that I don't want 2 in it even once. So, those divisors will be Q plus 1, R plus 1, S plus 1. If you add these two, then total number of divisors will be. Many equations are made in this category. We have a number N is equal to 3 power 6 multiplied by 5 power 9 multiplied by 7 power. Suppose this is 11. Now we have to find number of devices which are perfect square. If the device is perfect square, then how many times should this 3 factor come? Even number of times should come. In fact, each prime factor should be even number of times. So how many possibilities are there for 3? What can be done for 3? It can be done that 3 is not once or twice or 4 times or 6 times. There are 4 possibilities for 3. What are the possibilities for 5? 0, 2, 4, 6, 8 and for 7, the possibilities are 0, 2, 4, 6, 8 and 10. So, the total number of possibilities will be here. 4 multiplied by 5 multiplied by 6. So, the answer will be 24 multiplied by 5. That is 120. So, 120 divisors will be perfect square. Okay, let's say we take a number, number n is equal to 3 power 7 multiplied by 5 power 8 multiplied by 7 power 10. Let's take this number and you are told to find the divisors of type 4 lambda plus 1. We have to find out the divisors of this number. When we divide the divisor by 4, the remainder will be 1. The questions of number theory seem a bit strange. But if we think about it, then... So, how will I start? I will write a divisor of this number in general form, in which there will be some powers of 3, 5, 7. Say 3 power p multiplied by 5 power q multiplied by 7 power r. 4 lambda plus 1, I write 3, 4 minus 1 power p, 5, 4 plus 1 power q and 7, I write 8 minus 1 power r. Like this. Okay, so now this number 4 minus 1 power p, when I expand it binomially, then what type of number will it be? It will be 4 lambda 1 plus minus 1 power p. Now it will depend on p, whether it is odd or even. According to that, here plus 1 will come or minus 1 will come. This number will always be 4 lambda 2 plus 1 type, because it is plus 1. And again here minus 1 is there, so this number will be 4 lambda 3 here is 8 which is the multiple of 4 that is why I am writing 4 here 4 lambda 3 plus minus 1 power r now this number has to be of 4 lambda plus 1 type this is already in the form of 4 lambda 2 plus 1 so if we want to bring this product in this form then definitely these two values will be minus 1 or This means that for this number, the value of Q can be anything. So, in this 5 power 8, how many possibilities will be there of 5 powers? 9 possibilities. So, from 9 ways, powers of 5 can come here. The powers of 3 and R are connected to each other. It is a dependency. Both of them will be plus 1 when P and R will be even. What are the even possibilities of P? P means the power of 3 is 0, 2, 4 and 6. And what are the possibilities of its even? 0, 2, 4, 6, 8 and 10. Here are 4 possibilities. Here are 6 possibilities. So, what will happen? 24 possibilities for this to be even. Now, we will consider the cases of odd. When it is odd, then what is possible? 1, 3, 5, 7. And here, what are the possibilities for odd? 1, 3, 5, 7 and 9. So, from here, 4 possibilities, from here, 5 possibilities. So, total possibilities will be 20. So, either both are even or both are odd. So, this answer will be 9 multiplied by 44. So, these are number of divisors of this form. Okay, let's do a question which is a question which came in J 2010 in comprehension. The question is from matrices chapter but counting principle is to be applied here. Question is P is prime number and TP is set of matrices A, A, B, C, A. and ABC belongs to 0 to P minus 1 then we have to find number of matrices in this three cases case one hey question number one hey number of a if a symmetric or skew symmetric or both and determinant of a is divisible by P so Pella case later hey hum a is symmetric If A is a symmetric matrix, then definitely B is equal to C. So, at that time, the value of determinant will be determinant of A will be A square minus B square. This must be divisible by P. So, what we do is we make its factors. A minus B and A plus B. Okay. Okay. Okay, now this... If it is divisible by P, then one factor will be divisible by P, then our work will be done. If both are there, then also it is a good thing. Now we target A minus B. When will this be divisible by P? Now if you are taking a and b from here, then the value of a minus b will never be divisible by p. You can check it. If you take any two integers from here, then the difference will not be divisible by p. p is a prime number. So if I want to make it divisible by p, then a is equal to b. So how many values of a are possible? p values starting from 0 to p minus 1. So here we have p cases. A will be the same as B so there will be no chance of selecting B A will be equal to B and when A and B value will be 0 at that time this matrix will become null matrix which is symmetric and skew symmetric both So, P cases will be when A minus B will be divisible by P. Now, let's target A plus B. If A plus B is divisible by P, then definitely its sum should be P. Again, P is prime number. So, how many cases are possible? Here we start from 1, we are not taking the case of 0 because the case of 0 has come here. If it is 1, then what will be b? p-1 If it is a2, then it will be b, p-2 Last, when a will be p-1, then what will be the value of b? The value of b will be 1. So how many cases are these? These are p-1 cases. So, p minus 1 cases are coming from dividing a plus b by p. p cases are coming from dividing a minus b by p. So, total cases will be p plus p minus 1 that is 2p minus 1. So, the answer will be 2p minus 1. Okay. Let's try the second part of this first question. question is number of a if trace of a is not divisible by p by the way trace of a diagonal principle diagonal elements sum trace of a will be 2a this is not divisible by p but determinant of a is divisible by p determinant of a in this case this is a square minus bc Now, this is not divisible by P. So, you can take anything other than 0, 1, 2, 3 up to P minus 1, this will never be divisible by P because P prime number will not be there. So, what we have to observe here is that the value of A should not be 0. So, A is not equal to 0. Rest, A can take any value from 1 to P minus 1. Now, let us come to this determinant. So, we can select the value of a from p minus 1 way. By the way, this should be divisible by p. So, we divide it by p. So, this is a square upon p minus bc upon p. Now, the value of a is from 1 to p minus 1. You are doing its square, then you are dividing it by p. So, the remainder that you will get here, its value will also be from 1 to p minus 1. Anything can happen. Obviously, when you divide a number by p, then the remainder will be less than that. It will be less than 1 to p minus 1. remainder can go and the same remainder which is coming here, that remainder should also come here. Now suppose if I select B independently, then for B how many possibilities do I have? Again P-1 possibilities are there because B can also be any number from 1 to P-1. I can't take B as 0 otherwise it will become 0 then I won't get the remainder as much as I get from this division so for B I have cases P-1 ok now you have selected B independently so how many possibilities will be left for C? only one possibility will be left because we have to divide this product by P whose remainder should be same as this division so it can be in one case only ok So, for C there is one possibility to select. If you select C independently in P-1 way, then there will be one possibility for B. It means that you can select A and B in P-1 way and C in 1 way. So, the total cases will be P-1. You will multiply P-1 with P-1. So, the answer to this question is... will be p minus 1 square. This is the case where a value is not 0. Third part is number of A if determinant of A is not divisible by P. So in this case we will take out the total number of matrices. From that we will minus those cases when determinant of A value will be 0. Total matrices will be ABC 3 numbers. For all of them P possibilities are there. So total number of matrices are PQ. Minus. So in this we take two cases when a is 0 and when a is not 0. Okay, when a not equal to 0, at that time how many... matrices will be formed whose determinant will be divisible by p. What did we get from the second question? p minus 1 square. And what do we take a when a is equal to 0? When a is equal to 0, then the determinant of a will be 0 minus bc. Now this has to be divisible by p. That means bc should be divisible by p. But because bc is coming from this list, This product is divisible by P when either of these two is zero or both are zero. So you can check when there is no zero in B and C. At B and C, If you are taking the number from 1 to p-1, then the product of that number will not have the prime factor p anywhere. If the prime number p does not come, then it will not be divisible by p. So, it is necessary for this product to be 0. Okay, now when b is 0, if b is 0, what are the possibilities for c? Suppose we take c as non-zero, then the possibilities of c will be p minus 1. And if c is 0, then the possibilities of b will be non-zero possibilities p minus 1. Total possibilities are 2p minus 2. And in that case, where b and c both will be 0, then plus 1, then this will be 2p minus 1. So, these are the number of cases when a is equal to 0, we will also subtract this. So, your answer will come here. This is p cube minus, this is p square minus 2p plus 1 plus 2p minus 1. This is cancelled. So, the answer is p cube minus p square. So, see this equation is not as much than the matrix, as it is from the counting principle. Let's try a question which came in JEE 2009. The question was that you have a matrix of 3 x 3 and this matrix is symmetric. and its elements are 0, 0, 0, 0, 1, 1, 1, 1, 1 0 is 4 times, 1 is 5 times this matrix is made up of 9 entries and it should be symmetric I have to tell you how many such matrices can be made this was also a question from comprehension this is a part of it Here we make a principle diagonal and these are symmetric matrix so the elements in front should match like this. These two will match, these two will match and these two will match. Okay? Where to start? Now this is confirmed. What do we call this? This is called upper triangle and this is called lower triangle. So, this is called upper triangle and this is called lower triangle. Lower triangle should have the same number of zeros as the upper triangle. The number of 1s in lower triangle should be the same number as the number of 1s in upper triangle. That means the entries of 0 and 1 in upper and lower triangles should be of even number. So keeping that in mind, first we understand the diagonal. Can all three elements of the diagonal be zero? It is not possible because if you use 0 here, then only 0 will be left. So it will not be symmetric. So this is not possible. It is possible that two zeros and one 1 are there. I am talking about diagonal here. Okay, so we freeze this in diagonal, 001. Now, how many ways can these elements in diagonal come from? They can come from 3 factorial upon 2 factorial, that is 3 ways. Simply, 001, 010 and 100, from there. Now what elements will we have left? The elements we have left will be 0, 0, 1, 1, 1, 1. 4 1s and 2 0s. So definitely in this upper triangle, 1 0 will come and 2 1s will come. And 1 0 and 2 1s will come here in the lower triangle. Okay, now we have three places for 0, 1, 1. How can we arrange these three objects on these three places? Three factorial upon two factorial, that is three ways. When you arrange 0, 1, 1 here, then you have only one way to fill these elements. Because the element that will be here, will come exactly the same element here. So, here the total number of cases will be 3 multiplied by 3. See, this is a... to fill up diagonal and this is to fill up upper triangle and lower triangle so we have this nine matrices okay now let's come back to diagonal in diagonal this case is possible one zero and two ones no otherwise we will have three zeros left which are odd numbers their distribution will not be able to happen here so this case is not possible Then comes all 1. This is possible. At that time we will have entries 2 1s and 4 0s. And what will be its number of ways for upper triangle over triangle? 3 factorial upon 2 factorial that is 3. And this case is... almost the same because when we have these entries, then what can come inside the upper triangle? 1, 1 and 0. So, the number of those will be 3 only. So, our answer will be total 9 plus 3. There will be 12 matrices. Okay, now let's understand how many common equations which are there in books with JEE equations. match with concept point of view like the question is there are four pair of shoes and you are selecting four shoes in such a way that no pair is selected there are four pair of shoes and no pair is selected all the shoes will have one left and one right so we will write it like this Now you have to select 4 shoes, no pair is being selected. So definitely what will come out of every pair? One shoe will come. How many ways you can select 1 from this 2? 2 ways, from here also 2 ways, this 2 ways and this is 2 ways. So what will be the answer? 2 power 4 that is 16. But if in the question, instead of 4 pairs, there are 10 pairs of shoes. And you have to select 4 shoes in such a way that no pair is selected. So, I will write the pairs here. L1, R1, L2, R2, L3, R3, up to L9, R9, L10, R10. You have to select 4 shoes which will definitely come from different pairs. But the problem is which will be those 4 different pairs. So our first task is to select 4 pairs from these 10 pairs From which 1 pair will be the shoe So we can select those 4 pairs in 10 C 4 ways Suppose you have selected L1 R1, L2 R2, L9 R9 and L10 R10 Now, from these selected 4 pairs, what will happen is that from each pair, 1 shoe will come. So, how will we multiply? 2 x 2 x 2. From each pair, there are 2 possibilities to select 1 shoe. So, the answer will be 10C4 x 2 power 4. Now, the similar question came in GE. Let's see that. Okay, the question came in GE Advanced. In 2013, in integer type, the question was that V is a set of vectors a i plus b j plus c k such that a, b, c belongs to minus 1 to 1. Now three vectors are selected from this set. how many ways we can select if these three vectors are non-coplanar? First of all, let's check how many vectors we can make. A has two possibilities, B has two possibilities, C has two possibilities. So, total number vectors are 2 multiplied by 2 multiplied by 2. We have 8 vectors. Now, from these 8 vectors, we have to select 3 vectors so that they are not So, first of all we will write vectors. Vectors will be 1, 1, 1. Means these are the coefficients of i, j, k. In front of that we will write a vector. Minus 1, minus 1, minus 1. Whatever vector will come here, an anti-parallel vector will be created here. Because the possible values of our a, b, c are like this. What is one possibility? 1, 1, minus 1 So what will be its negative? minus 1, minus 1, 1 One will be 1, minus 1, 1 Its antiparallel minus 1, 1, minus 1 And last minus 1, 1, 1 Its antiparallel will be 1, minus 1, minus 1 So see, we have these 8 vectors made Now we have to select 3 non-coplanar vectors. Obviously, if you select 2 vectors from here, here, here, here, then the vectors will be coplanar. Because these 2 vectors are collinear. Similarly, these 2, these 2 and these 2 are collinear vectors. So if you select any other vector with these 2, then they will be coplanar. Because 2 of them are collinear. This means that we can select only one vector from here. In fact, we can select only one vector from any pair. But we have to select a total of three vectors. Now, we have to get one vector from each of them. So, what we will do first is, we will decide which three pairs of vectors we will select. So, we can select 3 pairs in 4C 3 ways. Suppose, you have decided these 3 pairs that we will select each vector from here. From here, how many ways can one vector come? From 2 ways. Similarly, from each pair, one vector can come from 2 ways. 4C 3 is 4 multiplied by this is 8. Answer is 32. So, we have 32 triplets of vectors which are non-coplanar. There is one way to do this. First of all, we will find the total number of vectors. We can select 3 vectors out of 8 in 8C3 ways. We will minus the vectors which are coplanar. If there are coplanar vectors then definitely one of these four pairs will be selected. We have four pairs and we can select one pair in 4C1 way. I have written 4C1, whose selection is this? It is of one pair. That means we have selected two vectors which are collinear. Suppose you are selecting this pair. You have selected this pair. So now we have to select one more vector, from which one can be? From these 6, any one can be, so multiply by 6C1, 1 pair and 1 vector, that means 3 vectors. 8C3's value will be 8 multiplied by 7 multiplied by 6 upon 6 minus 4 multiplied by 6, 24. So this is 56 minus 24, the answer from here is 32.