 Hi, I'm Zor. Welcome to Unizor education. This lecture is part of the whole course of advanced mathematics for teenagers, which you can find on Unizor.com. So all these YouTube lectures are linked into into this course with notes, with exams for registered students, and the course is free by the way. So everybody is welcome. Another very important thing is there are some problems, and in this particular lecture there are problems as well on combinatorics. And I always encourage people to go to the website Unizor.com, go through the notes, which contain basically the description of the problems, and try to solve them yourselves. Regardless of the result, whether you are succeeded or not, only then listen to the lecture and basically compare your approach to the problem, to the one which I presented. Now this is the beginning of the second part of the problems in combinatorics. The first part contains, if I'm not mistaken, seven different sessions. And I start this one from 2.1. I don't know how many sessions there will be, but I love problems in combinatorics because they are extremely suited to develop your combinatorial brain. And again, the development of your mind is the most important goal of this course, not something which you can just remember and use in your practical life. Well, speaking about practical life, this lecture is about combinatorics. And my examples are all from the game of poker. So you can consider this as part of the real life because sometimes people do gamble. Now, there are certain numbers which are presented in this particular lecture. These are basically relative frequency of occurring certain combinations in poker. And I do believe that you shouldn't really sit at the table without having some notion about which combinations are more frequently occurring, which one less frequently. So that would give you the good idea, actually, whether you can succeed or not. So, OK, so let's just go to the problems. And again, as I was saying, these are all problems related to the game of poker. And the game which I am talking about is we have 52 cards of four different suits, 13 different ranks of four suits, 13 ranks within each suit, from two to ace. Two, three, four, five, six, seven, eight, nine, ten, Jack, Queen, King and Ace. Now, out of these 52 combinations, each player gets five, five cards. And the combination can include different, basically, there are different combinations which can be dealt to a particular person. And I'm talking about five different combinations in this particular session, five different problems which I'm going to consider. So these are problems about how frequently certain combinations occur. Now, first of all, how many different combinations of five cards out of 52 you can have? Well, that's obviously the beginning of everything. This is the base. And obviously, you have a number of combinations of 52 by five as the answer, which is 52 times 51 times 50 times 49 times 48 divided by one, two, three, four, five, which happened to be 2,598,960. So, two and a half million different combinations of five cards you can get. Now, which combinations are considered to be like winning combinations or at least combinations which we can talk about in the game of poker? Well, there are certain number of them and five of them we will consider right now. Maybe we will consider some other combinations in another lecture. But this one is about five different combinations. So the number one is combination which is called four of a kind. Now, what's the examples of this combination? For instance, you have 10, 10, 10 and 10 in different suits. Every suit has 13 cards. So we have four suits. So I can have four different tens, let's say 10 of clubs, 10 of diamonds, 10 of hearts and 10 of spades and something which is completely different. So this is a combination when you have four of the same rank, four cards out of five of the same rank. Now, the question is how many different combinations out of this are these? All right, to count them, we have to actually make one particular decision. Well, the decision is what the rank of the combination which we are talking about. Well, we can have four twos, four threes, four tens, four kings, four aces. So we can have 13 different ranks which we can choose from. And then we can have all these cards of this rank represented. Now, so how many of them? Well, we can have 13 different quadruples of these cards, let's say four twos. We have right, so four threes, four fours, etc. So you have 13 different choices for this quadruple of numbers. So now with each of these 13 different combinations, we can have any one of the rest. Now, how many are left? Well, there are 52 cards, four we have already chosen, so there are 48 left. So each one of these can be with each one of these. So let's say four queens can be with three of diamonds or king of hearts or something like this, right? So we have this many combinations. This is the answer. Now, my notes usually contain the final result, final number. And during this lecture, I'll probably stop on something like this. I don't want to calculate the result. You can get it from the notes. But this is a number of different combinations which contain four of a kind. Next problem. Next problem is called full house combinations. Now full house is basically can be characterized with three cards of one rank and two cards of another rank. Okay, the question is how many of these combinations you can get? Okay, so first of all, we have to choose the rank of the three cards. Now there are 13 different ranks, so we have 13 choices. Okay, rank is chosen, but each rank contains four different cards, right? Clubs, diamonds, hearts and spades. We need only three of them. So now we have to basically multiply it by the number of combinations of three cards out of four. Now, first we chose the rank, which means we have chosen four different cards of the same rank. And then we have chosen a certain number of combinations out of this, which contain only three cards. So I have to pick three out of four. So with each choice of the rank, I have to choose three cards. Fine, three cards are chosen. Now the rest, the two cards, must be of a different rank, right? So I have to choose certain other rank from the rest, from the 12 ranks which remain. I have to choose one particular, right? So that's 12, 12 choices. Now, when this other rank is chosen, I need two cards from the four which are all of the same rank, right? So that means I have to multiply it by this. I have to choose two out of four. So this gives me the four cards of the same rank. This gives me only three cards out of these four. Now, this gives me one particular rank out of the remaining 12 ranks, which is also four cards, right? Four different suits. And this gives me a concrete two cards out of these four. So the multiplication of them, because with each of these, we can choose each of these, each of these, and each of these. The multiplication gives you the total number of full house combinations which you can get. And again, I'm not really calculating the result, it's in the notes. Next problem. Next is a combination which is called straight. Now, straight is a combination of five cards which have sequential ranks, for instance. Ten, nine, eight, seven, six, that's five cards, right? Or king, queen, jack, ten, and nine. Or ace, king, queen, jack, and ten. Or, and here is a very important, I would say, complication. The ace can be the very top card in front of the king. But it also can be on the bottom below two, which basically means that its value is one in this particular case. So this is also a combination which is a straight. So that's one complication. So ace can have both roles as the highest rank card and the lowest rank card. Another very important complication is that there is another combination which is also straight, but not just a straight, it's called straight flush, when not only the ranks are consecutive, but also the suits are the same. Now, but this is a completely different combination and it's called straight flush. We have to exclude straight flush from the straight. So when we're talking about straight, we're talking about five cards with consecutive ranks, but not of the same suit, like not all of them diamonds, right? Not all of them spades, etc. So we have to exclude these. All right, so how can we count the number of combinations of this kind? Well, first of all, let's count how many different top cards can be. Well, from five to six, seven, eight, etc. up to ace, right? So we cannot have four, three or two as the top card. So out of 13 ranks, three are basically cannot serve as the top, but the rest 10 can. So let's choose the top card and we have 10 different possibilities for this, right? Okay, great. Now, after that, we basically determine the rank of each card. So 10 is the rank of the top card, right? But that actually determines which means that the second card should be one rank below, the next should be one rank below, etc., etc., right? So it actually means that we have set, we have chosen the set of four cards for the next one. So it's nine of, for instance, if the first one is king, right? Then the second one must be a queen, right? So I have actually chosen four different choices for the queen and four different choices for Jack, four different choices for 10 and four different choices for nine. Choices means basically a suit which I'm basically choosing, right? And after I have chosen a particular rank, let's say king, I have also chosen four for different kings, right? Again, four different suits. So what I want to say is that as soon as I picked my one of the 10 different top cards, high-stranking cards, I have basically a choice of four on each one of these places, which means I have to multiply it by four by four by four by four, so four to the fifth degree, which is almost true except I have counted straight flush as part of the whole picture because they can all, in this particular case, can be all diamonds or all hearts, etc. And I have to exclude these straight flushes. So let's just calculate how many different straight flush combinations I have as a byproduct, basically, of this problem. Well, again, the first card can be any, which is 10, different choices, from five to eights. But then all other cards are predetermined. And the only thing which is not really free, we have to choose a particular suit for every other card, which is exactly the same as this one. So a pick of this card actually is equivalent to picking one particular suit out of four. So I have to multiply it by four. So I have 40 different straight flushes, different combination of the first highest ranking card and the suit. 10 different choices for the highest card, which is from five to eights, and four different suits. So as soon as I choose the top card and the suit, my combination is predetermined. So I have to subtract this from the number which I have counted before, and this is my answer. This is how many straight there are among all the combinations of five cards. So again, I have excluded these 40 straight flushes. But remember this, there are 40 straight flushes. Now next, well, next is a flush combination. Flush means we have all cards of the same suit. So you have five cards and they can happen to be, you know, all the same suits, all diamonds, all cards, etc. All right, so how can we find out this? Well, that's actually simple. First, you have to choose one of the four suits, which means you have four different combinations, right? Now that is a choice of 13 different cards of different ranks of the same suit, right? We have 13 cards of each suit. So as soon as we pick the suit, we have the 13 cards out of which we should pick five in our combination. Well, how many different ways we can pick five from 13? This how many, right? Now, what's wrong with this picture? Well, again, we have counted straight flushes because it happened, might happen actually, that all these cards which we have chosen, these five cards out of 13, they can have a consecutive ranks, which is not considered to be a flush. It's considered to be a straight flush, a completely separate combination. And we know there are 40 of them, right? So we have to subtract from here as well these which are straight flushes. So we subtracted straight flush from straight to get real straight, which is not straight flush. And we have subtracted 40 from the flush again to get out of the count straight flushes, which are not really just a flush. So this is the answer for the flush. And the last problem is three of a kind. So three of a kind. Now, you remember the first problem was four of a kind, right? When there are four cards of the same rank among our five. Now, obviously, the fifth card was supposed to be of some other rank because there are only four cards of the same rank, four different suits, right? Now, when we're talking about three, that's slightly differently, right? Three is different from four because if we pick three by accident among other two, we might actually get the same rank. So we have to somehow count it off. All right, so let's address the problem of three of a kind. All right, first of all, let's choose the rank of the three cards, which are supposed to be among these five. Well, there are 13 choices for the rank from two to eights. Okay, once I have chosen a particular rank, I have four different cards in that rank. Now, I need only three of them, right? So I actually have to multiply it by number of choices of three cards out of four. This gives me one particular rank. There are four cards in this rank, but I need only three. Now, all other cards are supposed to be not one of these ranks and they're also not supposed to be of the same rank because if they are of the same rank, I would have three plus two. I would have full house, which is a different combination. So somehow I have to take into account the fact that the other two cards are not supposed to be of the same rank and they're not supposed to be of this rank, which I have chosen. So to avoid four of a kind and to avoid full house, how can it be done? Well, very easily, they must be of different rank. So let's just choose two different ranks from the remaining 12 ranks by two. So what does it give me? Well, it gives me a combination of two ranks, not equal among themselves, which is different from the one which I have chosen the first because I have 12 here, so from the rest of the different ranks. So if this is, for instance, chosen as a king, now this is everything but king. And from everything but king, I have to choose two, for instance, ten and queen, two different ones. So that's why I have a combination. Now, once I have chosen that, I have two concrete ranks, which I have chosen. Each one of these ranks have four different cards of four different suits and I can have any one of the suits for one of these and any one of these and any one of the suits for another. So it's four choices for the fourth card and four choices for the fifth card. So I have to multiply it by four and by four. So that's the answer. So that's how many three of a kind combinations are among all the combinations of five cards. Well, that's it. That's my five problems. And I have considered five different combinations. There are some others, so maybe I will address them some other way. I would like to actually point out an interesting detail which probably will be related to my probability course, which I'm going to continue after the combinatorics. I'll go to probabilities. And I will basically compare the probabilities of different combinations. And that would be, well, the basis for decision making when you are playing the game of poker. All right, that's it for today. I recommend you to go through these problems yourself. The problems are in the notes in Unisor.com. And well, compare it with the answers. Answers are provided. So I think it would be very beneficial if you can just go through these problems yourself and come up with the same answers. And if you come up with some other solution which leads to the same correct answer, by all means, you can send it to me and I can publish it on the website with proper attribution to you as authors. So that's it. Thank you very much and good luck.