Before we move on with combinatorics, let's have some practice. Let's start with the following problem. Suppose we have a standard deck of 52 cards, and suppose you want to compute the number of all possible five card hands. Here, is an example of a hand. Note that the order of cards in your hand is not important. So let's assume that your hand is unordered. So we pick a subset of five cards among 52 cards, and we know the answer is 52, choose 5, it is equal to more than two-and-a-half millions. Let's move on to the next problem. The second is similar, but now we want to count the number of hands in which two cards are hearts and three are spades. So here is our hand. Note, to get this hand, we need to pick two cards from the set of all hearts, and there are 13 of them. We need to pick three cards from the set of spades, and there are 13 of them again. To pick two cards, we have 13 choose 2 options. For three cards, we have 13 choose 3 options. We have to multiply them by the rule of product, so the answer is the product of 13 choice 2 and 13 choice 3. It is more than 22,000. Now, let's proceed to the next example. Suppose now that we are considering four digit numbers, and I would like to compute how many numbers that have at least one digit seven. Let's try to do it. It turns out with it is not very clear how to do this. For example, we can start just by trying all possibilities for all digits. If we try all possibilities for the first digits, there are 10 of them. There are 10 possibilities for the second, 10 possibilities for the third digit, but for the last digit, the number of possibilities depends on wherever one of the previous digits was equal to seven. So it is not clear how to compute this number in this way. But we can use another approach. It turns out that it is very easy to compute the opposite. Let's compute all numbers that doesn't have digit seven. There are 9_4 of them. The reason is very simple, for each digit to have nine options, and by the rule of product, we have to multiply them, and the answer is 9_4. So there are 9_4 numbers with no digit seven. There are 10_4 numbers in total. So the answer to our problem is 10_4 minus 9_4. Let's proceed to the next problem. How many four-digit numbers do we have such that their digits are decreasing? So each next digits is smaller than the previous one. Let's try to do it. If we try to count options for each position, then it is complicated. For the first position, there are 10 options, but for the second, the number of options already depends on the first number. So it's not very clear how to compute this. The idea here is to look from the oversight. Let's consider four positions in our number. What we need to do, we need to pick digits from 0-9 to be in our number. Note that once we picked four distinct digits, our number is uniquely determined. Why is this so? Let's consider an example. Suppose we picked three, four, two, seven. There are four distinct digits. Note that there is a unique way to write them down in our positions in such a way that the numbers in our positions decrease. We can write with solving the following way; seven, four, three, and two. So once we picked four digits, the number is uniquely determined, and the order of our picks doesn't matter. So we just actually pick combinations of size four from 10 options, from digits from 0-9. The answer is 10 choose four which is equal to 210. Let's move on to the last problem in this video. Let's consider a chess board and we have a piece in the left bottom corner in the position zero, zero and want to move it to a position three, five. How many ways do they have to do it? Recall that we have discussed that there are exactly eight moves needed to move the position from zero, zero to three, five. Note that three of them should be moves to the right, and five remaining move should be moves up. Moreover, any such combinations of three moves to the right, and five moves up is a valid way to get to the cell three, five. So what do we need to do? We have eight moves, and we need to pick three of them to be moves to the right. So we have combinations, you have to pick a subset of size three from a set of size eight. So the answer here is 8 choose 3, which is equal to 56. It turns out that in this problem, we can actually see a Pascal triangle. Let's consider our chessboard, and let's in each cell write the number of possible ways to get to the cell from the position zero, zero, from the left bottom corner. Then in the bottom row, you have to write one because there is only one way to get there, we can only move to the right. In the first column, for exactly the same reason, you should write one for each cell here, there is only one way to get here. For the those cell in the second column. The second row, there are two ways to get here. We can get here from the bottom and from the left. There is one way for each of these possibilities. Next, for this cell, we have three possibilities because we can get here from the bottom, and there is one path will be sought. We can get here from the left. There are two ways of sort by the rule some of there are three possibilities here. For the same reason, there are three possibilities for the cell in the second column and the third row. Now here, you have four possibilities. Again, you can get here from the bottom. There is one possibility of this type, and we can get here from the left. There are three possibilities obviously. Here, we will have six because we have to add three possibilities, and three possibilities and so on. Observe that we actually obtain here a Pascal triangle. In this video, we have started binomial coefficients extensively. It turned out the we have several mathematical and combinatorial interpretations. We also had some practice to apply our knowledge. In the next lesson, we will see one more standard combinatorial setting.