 Hi, I'm Zor. Welcome to a new Zor education. I'm going to spend this time, this lecture, on solving certain problems related to cards, playing cards. Well, playing cards actually is a great tool for combinatorics and theory of probabilities. So anyway, I have four different problems. Now, this lecture is part of the whole course of advanced mathematics presented on Unizor.com. And what I suggest you to do is just go to the website and try to solve these problems just yourself. Very important. And then listen to the lecture, regardless whether you will be able to solve it or not. Now, the website contains obviously the problems themselves, the answers, short answers, so you can check your solution if you have one. And then there is a full explanation of the whole solution. So you can read afterwards the solution which I'm presenting. Okay, so four problems and they are basically about more or less the same thing with just different numbers. And again, that's about playing cards. I will use these playing cards as a great tool. Now, the deck I'm talking about is standard 52 cards deck and it has four different suits. It has clubs, diamonds, hearts and spades. So four suits, 13 cards each suit. So that's what we have. Now, let's talk about the problems. The first problem is very easy. You have to pick up six cards out of the deck of 52 cards. Six cards. And the requirement is they all should have exactly the same suit. So it's either six clubs or six hearts or whatever. So the question is how many different sets of six cards or how many different combinations of six cards all of which belong to the same suit you can pick. Well, this is a very simple problem and to solve it you have to do basically two things. Number one, you have to choose which suit you would like to be represented among these six cards. Now, how many different ways to choose one suit out of four? Well, it's number of combinations from four to one. Well, it's four actually. Now, the next step is after you have chosen a particular suit, it means you have chosen a particular set of 13 cards which the whole suit actually consists of 13 cards in the deck, right? So now you have to pick six out of these 13. Now, with each suit you can pick six cards out of 13 the number of combinations of six out of 13 times, right? So basically if you want we can calculate it very easily. Now number of combinations from four to one, well this is actually four divided by one, right? And the number of combinations of six out of 13 is 13, 12, 11, 10, 9, 8 divided by 1, 2, 3, 4, 5, 6. Now, I'm not going to count whatever these, I'm not going to do all this arithmetic. I refer you to the site Unisor.com where in the notes for this lecture there is the calculated result of this. Okay, so this is a simple problem and this is a solution. Problem number two very slightly more complex. Exactly the same deck of cards, exactly the same requirement to pick six out of 52 again and in this case my requirement is well the first problem was all of them should be the same of the same suit. Now all of them should be no more than two suits, alright? So it can be one but it can be two. For instance it can be all clubs or it can be clubs and diamonds or it can be all hearts or it can be spades and diamonds, whatever. But among these six cards we should have no more than two different suits, okay? That's the requirement. Alright, so how can we approach it? Well, let's try to approach it in a similar manner to the previous task. Namely, what did I do first in the previous problem? First I chose the suit and then I chose the cards from this chosen suit. Let me do the same thing. I know that the cards should belong to no more than two suits. So let's do first let's choose two suits out of four. That can be done in this number of ways different. So it's four times three divided by one times two which is actually so there are six different pairs of suits. Well, you can actually count it like clubs and diamonds, clubs and hearts, clubs and spades. Okay, that's always clubs. Now diamonds with clubs already hearts and diamonds and spades and the last one is hearts and spades. So these are all six combinations of two suits out of six. Fine, so I have chosen two suits which means I have chosen 26 cards, right? 13 cards of each suit. So my next problem is to choose six cards out of these 26, right? So that can be done in this number of combinations, right? Number of combinations of six out of 26. Well, and it seems to be reasonable to just have the multiplication of these two numbers as the answer. First we chose the couple of suits and then we chose six cards out of this chosen set of 26 cards. Well, not so fast. There is one little problem. Among these combinations which I picked from the 26 cards, among these six, you can have certain combinations which contain exactly two suits, like certain number of diamonds and certain number of clubs. And both numbers are greater than zero. Or you can have combinations where only one suit is represented. Same as in the problem one, like all six clubs for instance. Now, when I'm choosing six out of 26, obviously there are certain combinations which contain only one particular suit, one or another, right? Now let's think about it. You have, let's say, two different combinations of two suits, clubs and diamonds. Right? Now you also have clubs and hearts combination and you also have clubs and space combination. Now out of these 26 cards, I have been able to choose certain number of combinations which contain only clubs, right? But exactly the same combinations which contain only clubs, I can choose from these two suits, from these 26 cards. I count them separately, right? I'll just multiply it. Now if I'm multiplying, basically everything whatever is true for one pair of suits is true for another. And unfortunately I have counted my combinations which contain all six clubs here and exactly the same combinations from here and exactly the same combinations from here. So it looks like the same combination of six clubs I have chosen from these 26, from these 26 and from these 26. So I counted it three times instead of one. And this is true for any combination which contain only one single suit. Either it's combinations of clubs or diamonds or anything like that. So any combination which contain only six cards of the same suit, these particular combinations, I counted three times. Or other combinations which contain both suits represented among these six. I counted correctly once because the pair actually defines this type of combination. If both are present then obviously any of these combinations which contain both suits is different from this one. But if it contains only one combination then these three can be used to choose exactly the same six clubs. So I have to subtract from this twice because I counted three times so I have to subtract twice. I have to subtract twice the number of combinations which I calculated for the previous problem, the number of combinations which contain only one single suit. And this was C for 4 to 1, C 13 to 6. So this is the correct answer. This correction to my initial number was necessary because we have counted three times instead of once. Combinations contain only single suit. And this is the number which we have calculated in the previous problem. So that's the answer. So we are gradually increasing the complexity of our problem. The second problem is very much like the first one. Instead of one suit we just used two suits. But there is this very important nuance that we have counted three times something which we have to count only once. Alright, let's go into a little bit more complex case. We have three suits to be represented among the six cards. No more than three suits. So it can be all single suit or it can be two different suits or it can be three suits. So no more than three suits are represented among the six cards. Question is again how many combinations are there? Well, let's approach again the same way. First we can just start from picking three particular suits out of four which is four, three, two, one, two, three, which is four. But obviously it's the same as picking one suit because picking three means you are leaving one aside. So there are four combinations to leave one suit. So it's four combinations to pick three suits, right? That's kind of a symmetry. This is an obvious equality because both of them equal to m plus n factorial divided by m factorial and m plus n minus n and n factorial, right? And same thing here. So both are the same. So that's why number of combinations from four by three is the same as from four by one which is four. Okay, fine. So this is done. We have chosen three different suits which means 39 cards. So next step is from 39 cards we should pick six. Okay, now we know that this thing has certain problem. Something we have counted twice or three times or whatever. So let's just think about. Among these combinations which I have picked, six cards, there are some combinations which contain all three suits and these are fine because no matter which triplet of suits we have chosen, if my combination includes all three of them then all of them are different, obviously, with different choices. So that's fine. Now let's consider separately again those combinations which contain only one suit out of the three chosen or only two suits out of the three chosen. Okay, now the number of combinations with one suit. Okay, now let's say we have chosen clubs, diamonds and hearts. What other triplets are there? These are which contain clubs. So clubs can be a member of different combinations of suits. So among these 39 I have chosen a certain number of combinations which contain only clubs and exactly the same combinations I have chosen from these 39 cards and from these. So I have counted three times my single suit combinations and I have to subtract, obviously, two of them out. So my first subtraction is I have to subtract two times number of combinations of single suits. Okay, now let's talk about combinations of six cards which contain exactly two suits. Well, I have not calculated this number but let's just think about it this way. Which combination contains, let's say clubs and diamonds. Now what can be other variations of triplets? Well, it can be with hearts or it can be with space. So we have two combinations of three suits where a pair is repeated. Well, obviously if there are four suits, two I have fixed then I have picked the third one either one or another from those which are remaining. So my pair is contained in two different triplets. Which means that every combination which has exactly, not no more but exactly two suits is counted twice and we have to subtract once out of this quantity. But I have not actually calculated the exact number of combinations with two suits. What I did I have calculated the number of combinations with no more than two suits and separately the number of combinations with one. So if I will subtract from no more than two I will subtract all those which are exactly one suit combinations. I will have the number of combinations which have exactly two suits, right? So let's just think about it. I subtract a difference between number of combinations which contain no more than two suits which I have calculated in a previous problem minus the combinations which are exactly one suit, which is this one, equals two. Now this comes with minus. Now minus and minus and minus. So these are the same. This is minus three, right? So it's plus three, minus two, plus C4, 1, C13, 6. So that's the answer. So this is the number of combinations of no more than three suits. Which means among these six cards can be either one or two, but no more than three representatives of different suits. Alright. So that's the answer to this problem for three suits. And by the way, if you want exactly three suits, we have to subtract this no more than three suits. We have to subtract from this number the number which we had in the previous problem number two, no more than two suits, right? If no more than three minus, no more than two, I have exactly three suits, right? Okay. And the last problem would be, as I'm sure you have already guessed, four suits. Okay. First of all, it doesn't mean much to say no more than four suits because there are only four suits, right? And it doesn't mean much to say less than four suits because less than four means three and we have already covered this, right? So right now we're talking about exact for exactly four suits. So how many different combinations of six cards exist with exactly all four suits represented? So at least one card from each suit should be present among these six. And I'm going to solve this problem in two different ways. I think it's very important I told you many times before that it's important to be able to solve the problem from two different approaches. And the most typical way is one approach is directly count what's necessary and another approach count something which definitely does not belong to this category of necessary, but it's part of something common and just subtract from one another. So inclusion or exclusion. You can count what's included or you can have a total count and subtract what's excluded from the rule. So in this case, let me just do it both ways. And by the way, in the previous case I kind of alluded to the exclusion method. So I took the whole number of combinations and subtracted those which I have counted twice or twice or something. Now, how can I directly count the number of six cards combinations which contain all four suits? Exactly four suits. So it's one card from each suit at least. Well, let's just think about it. If I have six cards, I have at least one club, at least one diamond, at least one heart, and at least one spade, and I have two vacant places. Now, these two can belong either to the same suit, one of these, in which case I will have three cards of one suit, these two and one of these, and one card of the rest. So let's call it the combination three plus one plus one plus one, six cards. So three cards of one suit and one card from each of the remaining suits. Or they can belong to two different suits, in which case I have some of them will be two, another will be two, and then one and one, also six. There are no other cases. There are no other ways to put something, some suit belongness to these two, right? Either they're both the same or they're different, in which case we have either three one one one or two two one one combinations. So let's just count them separately and then add the results, right? That's reasonable. Alright. Now, how many different combinations exist of this type? Well, obviously I have to choose one particular suit which contains three cards and that can be done using this number of combinations. So I picked three card suit, one particular suit. Now, from this particular suit I have to pick sorry, I have to pick one suit out of four and this is the suit with three cards, right? And then from this particular suit I have to pick out of the 13 cards of this suit. I have to pick up three cards, right? Now, so I pick the large suit and this is one out of four and next from this suit I pick three cards. From all other suits, other three suits, each of them contains 13 cards. I have to pick one with multiplication obviously. And this is basically the number of all the combinations of this type. Again, we pick the cards the suit actually from which we will pick up three cards which automatically picks the rest obviously. And now I have four predetermined suits. One from which I have to pick up three cards and others I have to pick one. Now, let's talk about the second case. I have to pick up two suits from which I will choose two cards each, right? So now my suits have been determined. One suit with two cards, another suit with two cards and then one and one. Okay, so with two cards from 13 cards of the first suit I can do it this way. From the second one this way, from the third I have to pick only one card and from the fourth I have to pick only one card. And if I will add them up I will get the result. That's my direct calculation. Again, I'm not doing all the calculations. The calculations are done on the website unizor.com among the notes and if you want you can check your own calculations. Okay, this is the first method, the direct method. Now, let's try to do something similar to whatever I did before with the previous problems. Let's try to exclude it. But let's think about it. How many different combinations of six cards out of 52 we have? That's how many. Now, how many of them we don't want? We don't want those combinations which contain no more than three suits, right? So all the combinations which have three or two or one suit, we don't want. We want only four of them, all four suits represented. We have to basically subtract the result of problem three. Whatever the formula is. And here is an important thing which I consider actually an achievement. If you do this and if you will do all the calculations down to the numbers with the calculator, you will get exactly the same answer as I just did with the direct logic. So this is basically an ultimate test on whatever all these calculations are about, whether our logic is correct, whether we have subtracted, added, or whatever else we did. It's very, very important to have some kind of assurance that whatever you did is correct. So the equality between this number and the one before with 3 plus 111 or 2211. The difference between these numbers is basically assurance that all four problems are correct. Well, I wish you do it yourself. Just once more, just by yourself. Forget about whatever I said in the lecture. Try to remember it. Try to remember the logic. Go through the notes for unizord.com for this particular lecture. Well, check out your answer with the calculator. It's very important to get it from the beginning to the end with this particular correction checking whatever it's needed and get the right results. Well, by the way, if you really are comfortable with these four problems, which I have presented today, then I'm pretty sure you're comfortable with combinatorics in general. But I will continue solving certain problems. Now, this is the last lecture of the first part of the problems. Then I will have the second part. And by the way, I just want to mention one more thing, this DECO 52 cards. It's an extremely important tool for combinatorics. I said it before and I just love all these problems because we can just calculate certain things related to the games like poker, bridge, etc. And that probably would be a very good exercise for an inquisitive mind. Well, that's it for today. Thank you very much and good luck.