 So, we will be talking about the course on probability theory and its applications. Somehow, when you utter the word probability to a layman, it does not sound very familiar or you know one does not feel very comfortable with it, but my attempt in this course would be that at the end of the course you feel that you can understand what all goes behind computing the probability of an event. So, basically, probability theory is estimating the possibility of outcome of an event. So, this word possibility that what are the, how possible, how probable the event is, is actually done by counting and so therefore, before I start giving you axioms of probability theory, I would like to begin with the basic concepts of counting because that helps you in estimating the possibility of the occurrence of an event and I will try to explain what we mean by an event and so on. So, as we go on, hopefully you will understand all these terms. So, let me just begin why, so therefore, the first topic I am going to talk about is counting and I will start with an example. So, suppose there is a small community which consists of 12 women and each of whom has two children. Now, if one woman and one of her children, so it has to be the pair, a woman and her child. So, if the pair has to be chosen as mother and child of the year, how many different choices are possible? So, let us just, there is a very simple way of counting. So, what I am saying here is that first experiment would be number of possibilities of the, the first experiment would be choosing the mother out of the 12 women. We choose one of the women as the mother of the year and so the number of possible outcomes is 12 because they are 12 women and one of them will be appointed or chosen as the woman of the year and then we will, the second experiment would be selecting one of her children as the year, as the child of the year. So, once you have chosen the mother of the year, then she has two children. So, only one of them can qualify to be the year of the child, I mean child of the year. So, therefore, the possible outcomes are 2. So, then we say that the total number of choices is 24. So, I sort of broke up this event, I will call it event of choosing the mother and child of the year by first saying that I will choose the mother and then one of her children would be chosen as the year, child of the year. So, therefore, the number of choices are 24. So, if now I want to sort of formulate this into the basic principle of counting, then the basic principle says that if there are two experiments to be performed and experiment one results in any one of m possible outcomes. So, I have way of finding out that whatever my first experiment and the possible outcomes of that event are m. So, for example, here my first experiment was choosing the mother of the year and since there are 12 women out of which I have to choose the mother of the year. So, there were 12 possible outcomes of this experiment. Any of the 12 women could have been chosen as the mother of the year. So, the first experiment results in m possible outcomes. Now, for each outcome of experiment one that means for each possible out each of the m possible outcomes here, then I want to know what are the possible outcomes of the second experiment. So, in that example, the mother had two children. So, it could only be one of the two children who could be chosen as the mother of the year. So, now here we will say that if for each outcome of experiment one, there are n possible outcomes of experiment two, then together there are m n possible outcomes of the two experiments. So, I hope this once you understand this basic principle of computing of counting, then things become simple because you have to first find out what are the possible outcomes and then we will talk about when we define the concept of an event, then we will try to find out in how many ways a particular event can occur. So, here the total number of this thing are this. Now, if you want to generalize this concept of counting. So, this will be generalized basic principle of counting and here we will now say that suppose I have r experiments which could be 1 2 3 whatever the number and then again we will start saying first one the first experiment results in n 1 possible outcomes. So, the first experiment results in n 1 possible outcomes. Now, for each possible n 1 outcomes of the first experiment for each possible n 1 outcome of the first experiment, there are n 2 possible outcomes of the second experiment and then the third stage we will then count for each possible outcomes of the first two experiments because two experiments have taken place. So, for each possible outcome of the first and the second that means now you have n 1 possible outcomes of the first experiment, n 2 for each of the experiment outcome here you have n 2 possible outcomes. So, total number of outcomes become n 1 n 2. Now, for each possible outcome of n 1 n 2, the third experiment the possible outcomes will be n 3 and so on. So, you will go on counting this and therefore, by simple arithmetic you can see that the total number of possible outcomes would be n 1 into n 2 into n r that means at each for each experiment whatever the possible outcomes you will multiply them all. So, this gives you the generalized principle of computing. So, let us look up at this example, example 1.2 how many 8 place license plates are possible if first 4 places are to be occupied by letters and last 4 by numbers. That means the first 4 can be any of the alphabet A, B, C, D. So, there are 26 choices for the first 4 places and then for the last 4 have to be the numbers. So, which means they can be any of the digits 0, 1, 2 up to 9. So, now again as I quoted the generalized principle of counting. So, number of the first experiment would be choosing the first place of the license plate. So, then I have 26 choices because 26 alphabets are there. Then again for each of the alphabet I choose here for each of the 26 alphabets I choose here I can again choose the 26 alphabets again. If I have a letter A here for example, then I can choose any of the A, B, C, D, 26 alphabets here. So, again the next outcome that means the outcome for each outcome here there are again 26 choices of the second experiment and so on. So, again for the third that means now once have chosen the first 2 alphabets here then again for any each of these outcomes 26 into 26 I can again choose in the third place any of the 26 alphabets. So, then the third choices again the number of possible outcomes are the number of choices are 26 and the same argument goes for the fourth place. And similarly for the numbers I have chosen one of the 10 numbers here then again after having chosen all this I can again choose any of the 10 numbers here and then in the third place also any of the 10 numbers and so on. So, this will be the and you can just see I have left a question mark for you to compute the number it will be a big number. So, this many license plates can be there if you have this kind of arrangement that the first 4 places have to be occupied. So, it is a 8 letter license plate number and so the first 4 places have to be occupied by alphabets and the last 4 by digits. Now, if I change the experiment and I say that repetition of letters and numbers is not permitted. So, the moment I say that once I have chosen a letter here then the same letter will not be repeated here or here. So, then my way of counting will be this now for the first place I have any of the 26 choices any of the 26 letters can be chosen here. But once I have chosen a letter here then that same letter is not permitted to be chosen here. So, now my choice at for the second experiment for every possible outcome here goes down to 25 because whatever alphabet I have chosen here I cannot choose it again. So, therefore, my choices here are 25 and once I have chosen alphabet for the first place and the second place two alphabets have been selected then both these are not allowed to be chosen again. So, therefore, my choice for the third place would come down to 24. So, out of 26 two letters have been already chosen and they are not allowed to be chosen again. So, this would choice would be 24 and then similarly at the fourth place I will be having a choice of 23. So, you can if you just keep using this generalized principle of counting which I enumerated some time ago then you see this is how you will write down the possible outcomes. And similarly for the numbers for the first place I will have choice of 10, but once I choose a number here then my choice is limited to 9 and after this again I can only choose out of the 8 which are not repeated which have not been already chosen and similarly for the fourth place my choice will be left limited to 7 numbers which have not been repeated. So, this is the kind of so the generalized principle of counting helps you to come out with the possible number of outcomes of an experiment or event here what you want to say because this was as I am counting I am saying that this is 8 experiments choice of each letter for each place of the license plate and the choice of a number I counted as a experiment. So, I had 8 experiments here and use the generalized principle of counting to find out the possible number of outcomes. Now, another possible another way of counting is done by through permutations and combinations and let me now here try to explain the difference between permutations and combinations. So, here let us say we have collection of 3 novels and so I will just say that the authors are A, B and C then 2 mathematics books and again I will distinguish them or differentiate them by the authors and call the authors as D and E and one physics book and I will call the author as P. So, there are 6 books and since I am distinguishing the books by the authors. So, it is not just novels I am saying that the I mean the authors are A, B and C. So, there are different novels I am distinguishing similarly I am differentiating between the 2 mathematics novels and of course, there is one physics book. So, the question is how many arrangements are possible if the books are to be distinguished by the authors. So, I am wanting to make arrangements of these books and arrangements would be differentiated because I will be referring to the books by the authors. So, it is not just novels or a mathematics book. So, therefore, possible arrangements now here again let us just say that we are yeah right. So, here that means in the first place if we now count the thing then that means I can say that there are 6 places. So, now the choice for the first place that means I am just arranging the books like this in a row suppose. So, the choice for the first place would be any of the 6 books because I am differentiating the books by the authors. So, therefore, for the first place the choice is 6 books. Once a book is placed here then there will be 5 more left and again so therefore, out of the 6 experiments I am just saying that what are the possible outcomes. So, once I have chosen 2 books here then I am left with picking out the book for the third place from among the remaining 4 books and then 3, 2 and 1. So, the total number is 6 factorial which is 720. Now, the 720 arrangements are known as permutations of the 6 books and the permutation word is essentially saying that the permutations are the ordered arrangements of the books. Now, by order I mean that if I am writing say for example, ordered arrangement I want to say ordered arrangement. So, if I am writing here say it could be that I am right I have chosen all the 3 novels here right and then of course, let us say mathematics books you have D and E. So, maybe you have this and U right. Now, if I have the order B arrangement B A C E D P then you see this arrangement is different from this because now here the second author if you call the book by B as a second novel then the second novel is occupying the first place and this is first the novel by the first author is occupying the second place. So, here the arrangement and therefore, this order is also considered here in this arrangement and therefore, I will say that this is different from this order or this arrangement is different from this order that arrangement B A C E D P. So, this is the idea that your 720 equations are the ordered arrangements of the 6 books and so, how you place them depends on that means, I am distinguishing between which author is occupying the first place, which author is occupying the second place and so on. So, therefore, this way it will be 720, but if you do not distinguish between the authors that means, I just treat the 3 novels as novels and in that case this and this arrangement would be the same right and in fact, you can see how many possible arrangements see I can have here C A B E D P. Now, you can tell me 3 more because these 3 letters A B C themselves I can arrange in 3 factorial ways that will be 6 ways I can arrange these 3 and therefore, all those arrangements once if I do not consider the I do not distinguish between the 3 novels by the authors as long as they are just novels then that arrangement all those 6 arrangements will be counted as 1 arrangement. So, when we are looking at the arrangement of the books the novels 3 novels 2 maths books and 1 physics book. So, once we do not order we do not worry about the differentiating the books with respect to the authors in that case you see as I tried to show you that the 3 novels since they will not be distinguished by the authors. So, then for every arrangement for every arrangement in which only the arrangement of the the arrangement of the authors has been changed the other books remain the same in that case 6 of those arrangements will amount to 1 arrangement because the 3 novels can be arranged in 6 different ways and if I leave the maths books and the physics book intact then all 6 arrangements in which only the arrangements of the novels have been changed then those 6 arrangements will amount to 1 arrangement. And similarly corresponding to each of those if I leave everything intact and only disturb the arrangement of the maths books then since I am not differentiating between the authors those arrangements will not be different because the 2 maths books whether the author D or E whichever comes first does not matter to me. So, in that case I will then divide. So, that means when I am counting the arrangement where the order is not important then that comes out to be 6 factorial it was the arrangement total number of arrangements where the order was important, but if I do not care about the order or that means I do not distinguish between the authors in that case the arrangements will become 60. So, this is what now I am trying to come to. So, the in the permutations the order was important and then I said total number of permutation that means that the order of the objects was also being considered. So, now the general principle that I am trying to and see it that in general number of arrangements of n objects where n 1 are alike n 2 are alike and n are alike n r are alike. That means there are only are different kinds of objects and since. So, therefore, surely here n 1 plus n 2 plus n r adds up to n. So, you have n objects, but n 1 of them are the same then n 2 are alike and similarly n r of them are alike. And in that case when I am wanting to make arrange the n objects then the total number of objects would be because n factorial would be the arrangements of n objects when I am distinguishing between each of them right whatever the way of distinguishing the objects whatever it is. So, that was the total arrangement, but since as I tried to show you through this example of arranging the books that if n 1 are alike then those n 1 can be arranged in n 1 factorial ways and they will amount to the same arrangement because I am not differentiating between the order. So, therefore, I will divide this total number of arrangements here by n 1 factorial, n 2 factorial and n r factorial. So, that will give me the total number of so that becomes the number of combinations. That means the arrangements where the order is not important would be the number of total number of combinations. Now, let us look at this example. So, a tennis tournament has 9 competitors and then 3 from India, 2 from Japan and 4 from Malaysia. Now, if the results of the tournament are announced by nationalities of the players in the order in which they are placed. So, it is not you know you are not distinguishing the players by their names only their nationalities being considered. In that case 3 from India would be just 3 Indians and it does not matter which one gets the first position, third position or whatever it is. So, here it will be just all the 3 players would be considered as players from India. So, another way of looking at is that you can think of these 3 people as one entity because they are just representing the same country. So, similarly 2 from Japan would be just distinguished by their nationality and not by their names and from Malaysia. So, therefore, if you want to now find out the how many such lists are possible. That means, what is the arrangement of the position that these 3 players from 3 nationalities occupy in your list. Now, first, second, third, up to there will be 9 positions. So, then the total number of lists that you can have would be 9 factorial. So, 9 is the total number of players and if I was going to distinguish them by the names. In that case, there would have been 9 factorial arrangements of the way in which the order in which a players would occupy a position in the tournament, but since we are not considering the names we are only considering the nationalities. So, for example, we will then divide this number by 3 factorial, 4 factorial and 2 factorial to get the total number of lists. So, the 3 Indians can occupy any of the positions and they will be the same. So, that means, there are 6 lists in which if 3 Indians were occupying different positions. In the sense that I just rearrange the positions that are that means, 3 Indians if they just appear. See, suppose you consider the first position, second, third, fourth, fifth, sixth, seventh, eighth, ninth and same thing here. So, if this was Indian, this is an Indian and let us say this is an Indian. Now, whatever the 3 names, if I see if you had a name here is the Indian 1, Indian 2 and Indian 3, then I can have Indian 2 here, Indian 1 here, Indian 3 here and this way you can go on. Indian 2, Indian 3, Indian 1 and just see that you can write, you can rearrange these 3 in 6 possible ways. So, other arrangements will be in the same, but if these 3 I can arrange in 6 different ways, but for me they are the same because in my list it will appear as I, I, I. So, this will not be there. So, all these arrangements will be the same and so all these 6 arrangements will amount to the same. Therefore, I am dividing this number by 3 factorial and similarly for the 2 from I mean sorry here in this case here, that was the example for the books. So, here I am dividing this number by 3 factorial, 4 factorial and 2 factorial to get the possible number of lists. So, just trying to make the concept of counting clear, so that you can then you know when you have to compute the possibility of the occurrence of an event all these things will come in handy. So, now again I am trying to differentiate this principle that if you have a collection of n objects in how many ways can we select r objects that is again the same thing because you can say that you can put the n objects in a row and then you are picking up r out of them. So, that is a different arrangement there is another way of saying the same thing. So, here by a principle of counting when you have you can say that you have arranged all these n objects in a row and then you are picking up one of them. So, then for the first play that means when I want to pick up r objects from this list of n objects the first object there are n possible ways of selecting the first. So, we are counting the same thing again by a different way. So, here you have n ways of selecting the first object then since you have already selected the first object you are left with now n minus 1 ways of selecting the second object. And similarly, it goes on depleting and finally, for the r th object that you want to choose you are left with n minus r plus 1 because you have already selected r minus 1 objects. So, n minus r minus 1 this is the number. So, this many ways you can select the r th object. Now, since it does not matter again because I am not distinguishing between the r objects that means the order is not important. Once you see out of these n objects I have collected r objects. So, it does not matter which one came out first which one came out second as long as the set of objects is the same. So, that means my way of selecting may have been you know because I picked up the first object that counted as the first object second and so on. But then finally, what I have with me is the set of r objects and in that set the third or the fourth object could have been selected first does not matter. So, therefore, what we are saying is now you have n minus r plus 1. So, that means the total number I should have written the number here let me write somewhere here. So, the order is when the order is immaterial I am saying that this is let me just write it here. So, that means what I am saying is that the first object I could pick in any of the n ways then for the second object to pick up I had only n minus 1 choice is available with me and so on up to n minus r plus 1. So, this number and since what we are saying is that the it was not important the order in which the r objects were selected it is only the final subset of r objects that I have with me. So, therefore, I will divide this by r factorial and so you can write this number you see if you have up to this number you can write as n factorial divided by r factorial n minus r factorial. So, this is the final. So, that means this is the number of combinations I can have because by selecting r objects I can call an arrangement that was here that means the n objects that I have I pick up r I put them first and then the remaining n minus r. So, essentially the r objects that you pick it does not matter in what order you pick. So, therefore, the total number of ways in which I can arrange my n objects gets divided by r factorial and obviously, the last n minus r objects also the order is not important because finally, you have divided these n objects into r and n minus r. So, it is simply which r objects get selected it does not matter in what order they got selected. So, therefore, the total number of ways in which you can pick up r objects from n objects would again be given by this number and our normally we say n choose r this is the notation for this number n factorial divided by r factorial and minus r factorial. So, this is n choose r. So, again let us take up an example. So, a jury of 7 is to be formed from a group of 30 people how many different juries can be formed. So, just carry over the what we discussed just now. So, there are 30 people and you have to pick up 7 people from this set of 30 people which will form a jury and so you want to know how many the question is how many different juries can be formed. So, this is like we discussed n objects you want to pick up r objects from there and how many ways are there how many possible combinations are there that is what we are looking at. So, obviously, you can see that the number of juries would be. So, first person in the set can be chosen out of the 30 people then once you have chosen that person you are left with 29 choices then again after these two you are left with 28 and so on. So, this number is 24 and now what we are saying is that it was not at all important as to when we have selected the 7 people since this is the subset of 7 people it does not matter which one of them got selected first. So, the order is not important it is only that finally, I have this subset of 7 people. So, therefore, the 7 we will have to divide by 7 factorial to get the possible number of juries because this 7 set of people that you are forming the jury the you are not looking at when the first when the in what order they got selected that is not important. That is what I am emphasizing again and again that the here the order of selection is not important it is just that we want this subset of 7 people and therefore, this will be the total number of arrangements and so here with this number we can write because this is up to 24 remember this is n minus r plus 1. So, here your n is 30 and your r is 7. So, when you want to divide from 36 you are left with 24 and so this is the total number if it was important as to in what order the people got selected for the jury. Since it is not important it is not being considered it is immaterial. So, therefore, I divide by 7 factorial and this number can be written as 30 factorial divided by 7 factorial into 23 factorial which our by our notation is 30 choose 7. Now, in case the group of 30 consists of 10 women and 20 men and if it is required and if it is required that 2 women and 5 men should form the jury. So, now we are saying that out of the 10 women 2 must get selected and out of the 20 men 5 men should get selected to form the jury. So, in that case the number of groups of women that you can form out of you know by choosing 2 out of 10 this will be 10 choose 2. Here again the order is not important I just want to form select a subset of 2 people 2 women from out of the 10. So, that will be total number of ways in which I can select the 2 women from the 10 women and similarly I can 20 choose 5 is the number of ways in which I can select 5 men from the set of 20 men and so the total number would be because for each subset that I choose here of women there will be this many. So, I am now using here by generalized principle of counting and so the total number of ways in which I can the number of juries of 2 women and 5 men can be formed is this. So, just through examples I am trying to make the concept clear and now let us see that you can all of you have used binomial theorem and you know that if you want to expand this number x plus y raise to n then the formula is n choose k x raise to k y raise to n minus k and you have done it by induction and so on the proof. Now, let me give you this combinatorial proof of binomial theorem using this concept of counting that we have learnt here we can use this and show that the expansion of this term can be written in this way and so let me consider the product x 1 plus y 1 into x 2 plus y. So, the n terms here now when I am multiplying each term here final in the product would be consisting of n factors a factor of n of the x i's and y j's that means each term of this project product will contain when you count the number of x i's and y j's that total number would be n because you have n factors, but certainly in any factor when you look at the x i's and y j's if an x i appears then the same index will not be for y because you see x 1 and y 1 are in the same term here. So, therefore, when I multiply either I will be multiplying x 1 by the other terms and so then y 1 will not appear in that. So, therefore, in any product you the x i's and y j's that you have if an x i appears the corresponding y j y i will not appear in that product. So, in other words what am I saying that the n the any term of this product which will contain let us say k of the x i's and then the remaining n i minus k of the y j's would be such that. So, you are choosing essentially each term here would contain let us say when I am looking at the term containing only k of the x i's and so remaining n minus k are y j's. So, then how many ways can be there how many such products can be there in which. So, you know this is containing the each term in this product has n factors containing x i's and y j's. So, then if I am looking at all the terms which contain k of the x i's then that means I am out of the x 1 x 2 x n these n x i's I am choosing k of the x i's and then of course, the moment I choose my k x i's then the y j's got the n minus k y k's y j's got get selected automatically because the once which are not appearing in my x 1 x 2 the product of x i's that I have taken the remaining indices will be will go to the y j's right. So, here that means and that ways the way of choosing k x i's out of the n x i's x 1 x 2 x n is n n choose k right. So, therefore, the number of terms which contain k of the x i's is this many terms and. So, this whole product either what will happen see for example, if you look at the product y 1 y 2 y n no x i appears. So, either so the k can vary from 0 to n the terms will contain either 0 x i's then 1 x i 2 x i's and so on. So, you want to add up all such things. So, you want counting the total number of terms in this product and that total number will be given by summing up n choose k from 0 to n and when you choose like this then that means your k of the x i's are here and n minus k are here you are adding up right. So, this thing then I put x i equal to x for all i and y j equal to y for all j. So, in that case your k of the x i's that you took from here they all become equal to x to x raise to k y raise to n minus k. So, then therefore, this product and when 1 you put the x i's together equal y i's all together then this whole product coincides with this and so you have a nice way of you know proving the binomial theorem. Now, I can immediate application of this is that if you want to count you have n objects and you want to count how many subsets can be there of these n objects you know when of course, the empty set also we consider as a subset that means nothing gets chosen from the n objects. So, you can have subsets consisting the empty set then you can have subsets consisting of only one element from one object from these n objects subsets containing 2 objects 3 objects and so on. So, if you want to count the total number of subsets that you can form of these n objects then since I told you that n c k gives you the number of subsets of size k and so you want to count this number n c k from k varying from 0 to n and which if you look at this expansion here essentially you are putting x equal to 1 and y equal to 1. So, that this all becomes 1 and so you are summing up. So, on this side when you put x equal to 1 and y equal to 1 this reduces to 2 raise to n. So, the total number of subsets this is another nice way of counting the number of subsets that you can form given n objects. So, extending this concept that I just enumerated for you for the binomial theorem now we can take over go up to multinomial coefficients. So, this is a set of n distinct objects is to be divided into r distinct groups of sizes n 1 n 2 n r for the binomial theorem r was 2. So, now it is n 1 plus n 1 n 2 n r where of course, all the n i is add up to n. So, I am dividing these n objects into r subgroups and the size of each group the first subgroup is n 1 size of second group is n 2 and so on right. So, then using the same principle I want to choose from n n 1 possible groups. So, the number of possible groups of size n 1 is n n 1 and n choose n 1 right or you also write this as actually this is also written as n c n 1. So, combinations so n 1 combination that means you want to choose n 1 objects out of n. So, what are the possible number of ways. So, any of these notations is acceptable fine. So, the first group will be the number of groups of sizes n 1 would be n choose n 1 then since I have already chosen n 1 objects out of n. So, then I want to choose the second set of objects n 2 consisting of the set of objects must be n 2 in size and then so out of n minus n 1 I want to choose n 2. So, therefore, each group of size n 1 the number of choices of the second group is n minus n 1 choose n 2 right and so you extend this concept and so finally, the last choice would be because now you will be left with n minus n 1 minus n 2 minus n r minus 1 r minus 1 subsets or subgroups have already been chosen. So, then this many you are left and out of this you want to choose n r objects. So, again the number of possible ways is n minus n 1 n 2 minus n r minus 1 choose n r right and so we will say that the total number of groups is the product of all these right and so via notation this will be the first one would be n factorial divided by n 1 factorial n minus n 1 factorial then the second one would be n minus n 1 factorial divided by n 2 factorial n minus n 1 minus n 2 factorial and so on. So, if you see that this will be my total expression and then the terms cancel out because you have n minus n 1 factorial here and this is n minus n 1 factorial. So, it cancels out. So, these terms will cancel out you know in pairs and so in the denominator you will be left with n 1 factorial n 2 factorial n r n 3 factorial up to n r factorial and this is what again what we are saying is that we are deciding whatever some objects we are putting in one group we are saying they are all alike the same principle you are using and so their arrangements will not matter in whatever way you choose in whatever order you choose is in material. So, therefore, again I am I can arrange my possible objects n objects in n factorial ways, but since I am grouping them into r different subgroups and each group the number of objects is we are not differentiating between the objects in one group. So, then the total number of ways would be n factorial divided by n 1 factorial n 2 factorial n n r factorial and now this helps us to expand for example, this gives the multinomial theorem that if you want to expand x 1 plus x 2 plus x 3 plus x r raise to n then it will be because remember in this product just as I showed you for the binomial theorem you want to choose. So, because each product here will consist of some powers of x 1, some powers of x 2, some powers of x 3 and x r and so here I am now saying that these n 1 n 2 n r will vary from because every product here will contain 0 x 1 1 x 1 2 x 2 2 x 1 and so on and different. That means, whatever the powers each term here you can see in the fact in the product these terms will appear and so your n 1 plus n 2 plus n r must be because each term this is raise to n. So, each term in this expansion will contain n x i's together. So, that means you add up the indices of x 1 x 2 x r they should add up to n and that is what is given by this. So, therefore, using this concept this is how you can write the expansion and if you look at this example where x 1 plus x 2 plus x 3 raise to 3 then you see I can choose my n 1 to be 3. So, that means this subset contains just the x component the powers of x 1 powers of x 2 and x 3 are missing here. So, therefore, the expansion this is 3 3 0 0 x 1 raise to 3 then here again you are choosing a subset from 3 which consists only of powers of x 2. So, 0 3 0 and so this will be x 2 3. So, I showed you the idea behind this and then now I am using it repeatedly to say that how you can write the expansion of this. And so therefore, you can see that the way I can choose my numbers n 1 n 2 n r such that they add up to n. So, in the case when you have your r is 3 and your n is also 3. So, then these numbers must add up to 3 and you are dividing your number 3 in 3 possible ways. So, that they add up to 3. So, here these are the possible and the final thing is 3 1 1 1 and when each x 1 x 2 x 3 has only power 1. So, this is the expansion and now in the exercise sheet that we will just show you I am asking you how many terms are there in the in the multinomial expansion. And remember because which is actually counting the number of subsets right here essentially you are counting the number of subsets that you can have you know like you have r this thing here x 1 x 2 x 3 x r. So, essentially my question is how many terms are there in this expansion and so I have already discussed this case with you for the binomial how I used for answering the number of subsets of total number of subsets of n objects. So, here you have to use the same concept and tell me how many terms are there in this multinomial expansion. So, let me just show you the discuss the exercise sheet. Assign one is straightforward 18 workers are to be assigned to 18 different jobs. So, the important part here is that each job is different from the others and therefore, this will be an ordered arrangement of the 18 workers. So, you can write down how many possible assignments are possible that means an assignment would mean that you assign one particular worker to a particular job. So, then how many ways can you do this this is the idea here right. Now, consider a group of 25 people if everyone shakes hands with everyone else how many hands shakes take place. So, that means that a person shakes hands with another person then both have shaken hands with each other. So, just keep that in mind and you can immediately write down what the answer will be for number 2. Now, here in question 3 4 separate awards it can be you know somebody getting the highest marks the highest cumulative performance index best sportsmen leadership quality etcetera. So, you can have 4 different awards given to these students and the idea is to select from a class of 36 students who can be given this award. Now, in case a student can receive any number of awards then it will be a different number of arrangements that you can have or different ways in which the 4 awards can be given to a student or more than one student, but in number 1 the condition is that the student can receive any number of awards. So, you have to do a counting in that way in number 2 we say each student can receive at most one award. So, that means either a student receives an award or a student does not receive an award. So, this is the way you have to count. Now, in question 4 is interesting using combinatorial argument prove that n choose r that means selecting r items from n items is equal to selecting r minus 1 items from n minus 1 plus selecting r items from n minus 1 items. So, you can if you write down the expression for n minus 1 choose r minus 1 and plus n minus 1 choose r you can show that these 2 add up to n choose r right, but I want you to give a combinatorial argument and you can see here in the first term it says n minus 1 choose r minus 1 that means I am keeping away one particular object or item from the n that are there then I am selecting r minus 1. So, that means the first set of numbers n minus 1 choose r minus 1 gives you the number of ways in which you can pick up r minus 1 objects from n minus 1 when a particular object is been selected. So, when you add that to this selected group then it will become r objects from n objects. Now, the second one says that n minus 1 choose r that means that particular see the thing is that either the particular object or item that you have picked is either there in a collection of r objects or it is not there. So, the first case says that yes it is going to be in that collection of r objects. So, once you added to the set of r minus 1 objects that you have selected then it becomes r. Now, in the second one what you are saying is this is the set of this is the set of selections in which that particular object does not appear. So, you have separated that object then from the remaining n minus 1 you are choosing r. So, this is the kind of combinatorial argument you are giving to prove the identity. And so, you see I have already shown you that you can by combinatorial arguments you can give show you prove the binomial theorem. Now, similarly you can try to prove number of such identities through combinatorial arguments. Then question 5 we have already discussed in how many ways can r objects be selected from a set of n objects if the order of selection is also to be considered right. So, question 6 one delegate each from 10 countries that include delegates from India, Pakistan, Bangladesh and Sri Lanka are to be seated in a row. So, the arrangement has to be in a row. Now, delegates from India and Sri Lanka are to be seated next to each other and delegates from Bangladesh and Pakistan are not to be seated together. How many seating arrangements are possible? So, the idea here is that you know you have 10 different positions people are sitting in a row. And you want people delegates from India and Sri Lanka to be seated together right which means that I can treat those two as one person. In that case it will be 9 people then who have to be arranged because delegate from India and delegate from Sri Lanka have to be together. So, that means the total arrangements is 9 factorial, but since the people are sitting in a row that means the arrangement that first the Indian delegate is sitting and then the Sri Lankan or the Sri Lankan is delegate is sitting first and then the Indian. So, this will count as two different arrangements. And therefore, the total number of arrangements would be 2 into 9 factorial in which the delegates from India and Sri Lanka are together. Now, you do not want people delegates from Bangladesh and Pakistan to be sitting together. So, now again consider the situation when they are sitting together. So, we will subtract the number of. So, in that case if you now look at the arrangements in which delegates from India and Sri Lanka are together and delegates from Bangladesh and Pakistan are together then you know you will have 8 different positions to arrange because these two delegates will be together that means that they can be treated as one person. And therefore, it will be total number of arrangements would be 8 factorial, but then again you have to it can be 2 possible you know like as I told you the arrangement can be India, Sri Lanka or Sri Lanka India. Similarly, it can be Bangladesh, Pakistan, Pakistan, Bangladesh. So, therefore, essentially you will be subtracting from 2 into 9 factorial the number of arrangements in which all the 4 delegates from Bangladesh and Pakistan are together and delegates from India and Sri Lanka are together. So, that will be 2 square into 8 factorial. And so, you subtract this number from 2 into 9 factorial that should give you the required number of ways in which you can have the required arrangement. So, how many terms are there in the multinomial expansion of x 1 plus x 2 plus x r raise to n. So, here again I am wanting you to do this exercise. See the whole idea the way I explained to you this expansion was that you are essentially dividing your number n into r smaller numbers n 1 n 2 n r. So, then they sum up to n. So, in how many ways can you partition this set of n numbers into r subsets. So, that they all add up to n this is the whole idea. And so, you will see that from this identity x 1 plus x 2 plus x r raise to n you have to put all these x 1 x 2 x r equal to 1. And so, you get r raise to n is equal to the number of terms that appear in the expansion multinomial expansion of this term. Then finally, a total of 6 gifts are to be distributed among 9 children. So, that no child receives more than one gift. This is exactly the same as you know part 2 of 3. Because, here also we said that no student should get more than one award, should get at most one reward. And so, here also we are saying that a total of 6 gifts are to be distributed among 9 children. So, that no child receives more than one gift.