 Now, there's a reason we call this branch of mathematics combinatorics and not permutatorics. And that's because any problem of permutations can be answered by the fundamental counting principle. Permutations are easy. But what if we have a combination? This is more difficult, but here's a general strategy. First, we'll determine the corresponding number of permutations, and then we'll determine how many permutations correspond to any given combination. For example, two out of 12 people are chosen to go to a conference. How many different pairs can be chosen? So first, we'll compute the number of permutations. And so let's think about this. We're going to choose two people. So we have a first person that we choose, and a second person that we choose. Now, since there are 12 people, we have 12 choices for that first person. And because we've already chosen one of those people, there's only 11 choices for that second person. And so there's 132 permutations. However, our actual choice of two people to go to the conference is not a permutation. Picking A as the first person and B as the second person is the same as picking B as the first person and A as the second person. So this is a combination. So let's think about that. Consider the combination like A and B. This selection of two people actually corresponds to two permutations. We either picked A first, then B second, or we picked B first and then A second. And in either case, we got the group A and B. And a little bit of thought should convince you that there's no other way to select A and B to go to the conference other than to either pick A first, B second, or B first, then A second. And so what that means is this combination A and B actually corresponds to two permutations. And we can generalize this. Every combination corresponds to two permutations. And that means that two times the number of combinations will give us the number of permutations. But we know the number of permutations. And that allows us to determine the number of combinations. What do three people go? Well, we'll start again by finding our permutations if we're going to choose three people. We choose the first person, a second person, and a third person. So we have 12 choices for that first person, 11 for the second, and 10 for the third. And so there's 1,320 permutations. So again, let's consider any combination of people A, B, and C. This group could have been chosen by choosing A first, B second, C third. But we could have also chosen them by picking A first, C second, B third. Or maybe we chose B first, A second, and C third. And there's a few other ways we could have selected the group A, B, and C. And in fact we see that this combination, A, B, C, corresponds to six different permutations. And so that tells us that six times the number of combinations is equal to the number of permutations. We know the number of permutations. And so we can solve for the number of combinations. Now with two or three things in a combination, it's not too difficult to figure out how many permutations each combination corresponds to. But as the number of elements in a combination goes up, this list gets longer and longer. And so we might ask, is there a way we could have found how many ways we can form one combination without enumeration? In other words, without listing every single possibility. So let's think about this as a permutation problem. So again we can think about this as answering a sequence of questions. And so if we're going to get the group A, B, C, we have to decide who we're choosing first, second, and third. And so for the first person, we have three choices, A, B, or C. For the second person, we only have two choices. And for the third, we have the last person standing. And so there's three times two times one, six permutations that correspond to the single combination, A, B, C. So suppose you're choosing five books from a library of 30, how many different choices are possible? So again, we'll find our permutation. We'll pick a first, second, third, fourth, and fifth book. And we have 30 ways of choosing the first book, 29 ways of choosing the second book, and so on. And so that gives us a heck of a lot of permutation of five books. But many of these correspond to the same combination of books. So let's think about that. Consider any combination of books, A, B, C, D, E. So to get this combination of books, we had to pick one of them as the first, one of them as the second, and then a third, a fourth, and a fifth. And so there are five books we could have chosen first, four books we could have chosen second, three as third, two as fourth, and then the last book standing is the fifth choice. There's 120 permutations corresponding to this one particular combination. And so that says that the number of permutations is 120 times the number of combinations. And we can solve that to find the number of combinations. Now, you might notice that in permutation and combination problems, products of descending sequences of integers appear frequently enough that they warrant a special symbol. And so we introduce the following definition. If n is a whole number, an exclamation point, which we read as factorial, is the product of the whole numbers from 1 to n. And because it will be convenient to have this value, we define zero factorial to be 1. Now, ordinarily at this point, we drop down a whole bunch of formulas that you could calculate the number of combinations. And those formulas are useful, but it's worth pointing out they're rarely usable. And in particular, in most cases where you want to find the number of combinations, no simple formula is going to be usable. We'll take a look at that next.