 Hello guys, so the World Cup Cricut 2023 fever is on and all of us are rooting for Team India to win this time. It's been 12 long years since we won the last World Cup in 2011 and we all watched it under the leadership of people like Tony, Sachin, Buraj, Kohli and so on where we last picked up the trophy. And since then there are a lot of gaming apps also where you can choose your own Dream 11 team playing 11 and then win exciting prizes. So let us combine a little bit of mathematics and a lot of cricket. And then the objective of today's video is to help team India choose the winning team of 11 players out of the squad of 15 players. So the squad of 15 players was announced recently on 28th of September which contains 7 bets including captain who is Rohit Sharma. Well, we have two wicket keepers Ishan Kishan and Kail Rahu to choose from. We have all rounders as likes of Hardik Pandeyo is the vice captain and Sajad Eja and we have none other than Buraj Kohli. So let's get to the problem directly. In this video I'm going to ask three problems. One is going to be purely simple but mathematically in nature. The second is going to be more realistic for you guys also help team India choose the playing 11 which can win the World Cup. And the third problem is going to be more mathematical in nature who are more interested in mathematics. Alright, so let's get to the problem. The first problem is in how many ways we can choose 11 players for playing 11 out of the squad of 15 players. The concept that is being used is from the chapter of permutations and combinations which you study in 11 standard in India. And if you're preparing for combinatorics olympias then maybe even in the 9th or 10th grade. And the simple answer is the expression is 15 C11 which translates to choose 11 player out of 15. And the mathematical expression for this is 15 into 14 into 13 into 12 divided by 4 into 3 into 2 into 1. Well if you calculate it turns out to be 13 65 which is lots of combinations to choose from. But the good news is well it's purely this problem is purely mathematical in nature. It doesn't have anything to do with reality. Because in the realistic scenario there are certain players which will always play like captain will always play. Vice captain will always almost play unless there is some injury. Also you must play with a wicketkeeper also if Kohli is there Kohli will play. He is a star cricketer and all of you love to watch Kohli. This is a large number of combinations. Let us see a realistic problem. Alright so the next problem is we want to make a realistic team and the following are the constraints. And how many ways can we choose 11 players out of this part of 15 with the following constraint. The first constraint is the captain and vice captain always play. Well it's obvious that you will want to play with both of these players. Exactly one wicketkeeper plays. Now we have simplified the problem. Realistically strategy depends from captain to captain and from selectors to selectors. But this is one strategy which is fairly decent and which I have chosen and which I feel is optimum. So exactly one wicketkeeper plays out of two wicketkeepers that we have. Exactly three ballers are chosen out of the four ballers that we have. And exactly three all-rounders are chosen out of which well vice captain is also there. Now you can imagine why these constraints because you want to deliver a spell of 50 overs. With at least five ballers plus all-rounders and having just five is also not good because you may want a change of pace, a change of balling variations and so on. So just having five ballers plus all-rounders is also not a good idea. Just betting has always been the strength of team India. So mostly our captain is a batsman. Even the wicketkeepers are always batsmen and all-rounders some of the all-rounders are also really good batsmen. Like Officer Jadija who has helped us with a lot of things. So let us see in how many ways can we do this problem. First of all you can guess how many combinations will we have within these four constraints. It will definitely be less than 1365 let us see how many. The constraints are Rohit Sharma and Hardik Pandya always plays. Alright so two are selected nine more to go. The next constraint was that you have to choose exactly one wicketkeeper out of these two and let me keep writing those numbers. If you have to choose exactly one wicketkeeper out of two you can do it in two C one number of pace. Next you have to choose exactly three ballers out of these four. And out of these four ballers exactly three ballers can be chosen in four C three number of pace. So how many people have you chosen so far? Well one captain let me write this one C one vice captain let me write this one C. So if I add all the lower indices I have the number of players. So one plus one two plus one three plus three six. So far we have chosen six players out of eleven. Now they have said they have asked us to choose exactly three all-rounders. Now we have already chosen one all-rounder and out of this remaining three all-rounders we can only pick two that can be done in three C two number of pace. And how many other cricketers remain? Well three four five six seven eight other three cricketers remain and how many other people remain? They are all batsmen so these are four batsmen out of which you have to choose three. So which can be done in four C three base. And if you count these numbers one C one is one two C one is two so the number turns out to be two and two four. Into three into four which turns out to be ninety six. It is less than hundreds much much less than thirteen sixty five. But still within these constraints you have a large number of combinations with which you can. Now individually what you can do is you can try to pick three of your four favorite batsmen. You can pick three of your four favorite ballers and two of your three favorite all-rounders. And please write your dream team in the comments below and let us see how many times that dream team is chosen. Also if you want to add one more realistic constraint to it. That sir Kohli always plays. Then we can just change these batsmen to out of these three batsmen you have to pick two. So if I change it to three C two that number will turn out to be two into four into three into three. Which will be significantly smaller which is seventy two. Because you want Kohli to play. So please write your dream team in the comments section. Let me summarize each of these again. What was once even it was for captain. This was for vice-captain. This was for choosing one cricket keeper. You can choose three ballers out of these four. You can choose two all-rounders out of the remaining three. And this is for choosing the remaining pure batsmen who are like neither captain nor cricket keeper. So you can choose three batsmen. And the yellow case is when sir Kohli is always chosen. Okay. Out of the three batsmen Kohli is definitely chosen. All right. So I hope you had fun making your own dream team. And let's do one mathematically challenging problem. Those of you are interested in mathematics can try it. Else can have fun while playing the cricket. And go team India. Let's win the World Cup tournament. All right. So the mathematically challenging problem is in how many ways can we choose eleven players out of these fifteen with the following constraint? Well, the obvious choices captain and vice-captain always play. Rather than exactly I'm writing at least one wicket keeper plays out of the two. At least three ballers are chosen out of the four. The same time at least three all-rounders are also chosen. But number of ballers plus all-rounders is less than or equal to seven. It cannot be more than seven. Well, that makes things slightly more complicated. So let us please try the problem on your own first. And then look at the solution next. So let me summarize it for you. So let us make categories right. And those categories are. Suppose I talk about captain, then there is only one. And then if I talk about, well, what are the different categories? There is captain, there is vice-captain. Then let me write the wicket keepers. Then let me write the ballers. Let me write the all-rounders and let me write the batsmen. Now ballers, all-rounders are batsmen, I'm assuming. They are batsmen which are neither captain nor wicket keepers. Similarly, all-rounders which are not vice-captain. So each of them in numbers are one captain, one vice-captain, two wicket keepers, four ballers. Other than the vice-captain, we have three all-rounders. And other than the captain and the wicket keepers, we have four batsmen. And let me take all of these and let us see what can we choose and how many ways can we choose each of these. So the constraints for captain and vice-captain are already chosen, always chosen without fail. So what we want to do is we want to make different cases. Let us see how many cases will be. So let us iterate systematically and let us handle each condition one by one. So the first condition was based upon the wicket keepers. So see, captain and vice-captain are always chosen. So we can always write one and one for captain and vice-captain. Out of these two wicket keepers, you have to pick at least one. So let us make all the cases in which we choose one wicket keeper. Now what was the next constraint? The next constraint was at least three ballers are chosen. So let me start with choosing the minimum number of ballers, which is three. And at least three all-rounders are chosen, out of which Hardik Pandey has already chosen. So out of the remaining three all-rounders, let me choose two more all-rounders. So what I have done is I have taken minimum of all of these. And if you take minimum of all of these, you have one plus two plus three, four, five, six, seven, eight players. So automatically out of the remaining four batsmen, you have to choose three batsmen. That is case one. Similarly, let us make all such cases. And on the right-hand side, we are going to count the number of ways of selecting them. In case two, captain and vice-captain are always chosen. Let us keep wicket keepers at two, ballers at three. Let me increase the number of all-rounders to three. Can I do that? We have, you know, vice-captain who is all-rounder plus ballers plus all-rounder at seven. Can we choose seven all-rounders? Yes, we cannot choose more than seven, but we can choose seven. And then obviously remaining, if I increase one of the all-rounders, I will have to reduce one of the batsmen. That is my second case. What other cases can be made? If in case three, let me choose one keeper, sorry, one captain, one vice-captain, one wicket keeper. Now I think with three ballers, can we make any other combination? Either we can have three all-rounders. If I increase the number of all-rounders to, let's say, so we have three plus one. If I increase the number of ballers to four. Next, let me increase the number of ballers to four. Then let me start with, again, number of all-rounders to be two. So total number of ballers plus all-rounders is one vice-captain plus who is all-rounder plus four ballers plus two, so seven. So yeah, this case is already also valid. And here again, we will have two number of batsmen. Let's keep going. Similarly, if we talk about case four. Now with one wicket keeper, we have only these many cases. So next, what we'll have to do is, we'll have to increase the number of wicket keepers. So let's say we take both keepers, right? So in case four, what you can do is, you have captain, vice-captain. Let's make it two wicket keepers. And then minimum number of ballers is three. If you choose two all-rounders, then you will have to choose only two batsmen. Well, wicket keeper is also technically a batsman. So number of batsmen are considered in this case and this case. Similarly, if I talk about case five and in case five, what we are doing is, we are choosing keeper, or sorry, captain, vice-captain, two wicket keepers. If we choose three ballers and three all-rounders, remaining, you have to choose only one batsman out of four. And finally, one more case will be there in which you pick case six, one captain, vice-captain, two wicket keepers. Let us increase the number of ballers to four. Then you can only have two all-rounders because the vice-captain, pander, is already all-rounder. So he has seven all-rounders. You cannot have more than seven. And in this case, how many batsmen have you chosen? One plus one, two plus two, four plus four, eight plus two, ten. So only one batsman can be chosen. Now we have made all the cases. So you can simply count the number of ways of choosing these. So for these two, number of ways of choosing one out of one is always one. So let us not even count them. You can choose two wicket keepers out of one. Two C1 ways into four C3 ways into three C2 into four C3. And here we are going to use addition rules. So we are going to add all of these cases. For the next one, the number of cases is two C1 into four C3 into three C3 into four C2. Here the answer will be two C1 into four C4 into three C2 into four C2. Well, the number of ways of choosing two wicket keepers out of two is always one. So let us not write that again. In these cases, you only are picking ballers out on us in batsmen. So you can do that in four C3 into three C2 into four C2 ways. Then you can do it in four C3 into three C3 into four C1 ways. And finally, you can pick four baller out of four and four C4 ways. This is three C2 ways and this is four C1 ways. All right. Now if you add all of this, what do you get is so this whole number is two into four into three into four four fours are sixteen and sixteen six are ninety six plus this turns out to be three C3 is one four C2 is six six into four is twenty four twenty four into two is forty eight plus well four C4 is one and three into two is six six into six is thirty six plus four C3 is four into three into six twelve into six is seventy two All right, eighteen fours are seventy two plus what we have is four into four that is sixteen only and finally what we have is three into four that is twelve and if you add all of these eight plus six fourteen plus six twenty twenty two and eight thirty carry three nine plus three is twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen plus nine is how much twenty eight so total number of combinations answer to this challenging problem is two eight please write in the comments how many of you were able to get the correct answer with all the six cases I hope you had fun learning maths as well as playing cricket so well keep studying keep watching cricket but only for a limited amount of time and all the best team India until next time