 Most problems in commentatorics are actually permutations, where the number of choices is itself a combination. For example, a company is sending seven people to a conference, one of its seven vice presidents, two of the six people in marketing, and four of the fifteen people in sales. How many different groups can go? And so we can think about this as asking a sequence of questions, which of the vice presidents is going, which of the two people from marketing are going, and which of the four people from sales are going. Since we can't switch answers, so the vice president is not the four people from sales, this is actually a permutation. So that means we just have to figure out how many possible choices we have for the vice president, how many possible choices we have for the two people from marketing, how many choices we have for the four people from sales. Now, since there are seven vice presidents, we have seven possible choices for the vice president. How about those two people from marketing? We can ask a sequence of questions, who's the first person we choose, and who's the second person we choose. But since we can switch our answers without changing our results, this itself is a combination. And so first we count our permutations, there's six people in marketing, so there's six ways we can choose that first person, and five ways we can choose that second person. So there are 30 permutations. Now consider any combination of two people, A, B. There are two ways we could have picked that first person, and one way we could have picked the second person, so there's two permutations of the elements of this combination. And so two times the number of combinations gives us the number of permutations, and so that says that the number of combinations is 15. And so we have 15 ways of choosing two people from marketing. And similarly, our choice of four people from sales is also a combination. So again, first we count our permutations, we have a first, second, third, and fourth person to choose, and we can do that in a whole bunch of permutations. And again, let's consider any combination, A, B, C, D, of these four people. In order to get this group of four, we have to have chosen one of them first, another one second, another one third, and another one fourth, and there are 24 permutations of the elements of this combination, and so that tells us that 24 times the number of combinations gives us the number of permutations, and so we know that the number of combinations is 1,365. And so there are 7 times 15 times 1,365 ways to choose seven people from this group. Again, there are formulas for computing the number of combinations, and in most cases those formulas are useless. So let's get lunch. We can get a meal by choosing one of eight entrees, two of 12 side dishes, and one of six desserts. How many different meals are possible? So again, if we consider our choices as being selecting an entre, selecting a side, and selecting a dessert, this is a permutation problem. Well, there's eight entrees and six desserts, but we need to know how many ways we can select two sides. Now, let's think about this. In some cases a single combination, say mashed potatoes and green beans, corresponds to two permutations. We can pick mashed potatoes first, then green beans, or we can pick green beans, then mashed potatoes. But some of us like to double up on our sides, so in other cases a single combination, mashed potatoes, mashed potatoes, corresponds to one permutation only. Pick mashed potatoes first, then pick mashed potatoes again. We'll consider these scenarios separately. So first suppose our sides are different. There are 132 permutations. Now, having chosen two different sides, there are two permutations that correspond to the single combination, and so the number of combinations is 66. But if the two sides are the same, well, our first side we have 12 choices, and for our second we only have one choice, namely whatever we chose the first time, and so there's only 12 ways we can pick two sides to be the same. Having chosen the same side twice, there's only one permutation that corresponds to this combination, and so there's only 12 combinations that consist of the same side repeated. And so we have 66 different ways of choosing two different sides, and 12 more ways of choosing the same side twice, which gives us 78 ways to pick two sides. And so the number of permutations is going to be 8 entrees, times 78 choices of sides, times 6 desserts.