 So a combinatorial question is, how many n permutations are there of the multi-set? There are two easy cases. If n is greater than or equal to any of the n i, there are i to the n permutations. And if n is exactly equal to the sum of the n i, then there are. But if n is less than n i, for some of the i, then we might run out of some of the elements. Now, if we run out of some of the elements, we'd need to break the problem into a number of distinct cases. Now, remember, you can assume anything you want as long as you make it explicit. And so we'll assume things that will make our life easier. Since we want to find the number of n permutations, we can't actually change n. But we can change our multi-set so that n is the sum of the repetition numbers. For example, suppose we want to find the number of three permutations of the multi-set. Now we could have zero, one, or two a's. If we have zero a's, then we could have zero or one b. If we have zero b's as well, then all three elements must be chosen from the multi-set with zero a's, zero b's, and three c's. And since the size of the permutation is the sum of the repetition numbers, we can use our formula to calculate the number of a's. If we have one b, then we must have two c's, so all three elements must be chosen from the multi-set with still zero a's, one b and two c's. And we can do this in. Now, since we only have one b in our set, we can't have two b's, so the next step might be to consider if we have one a. Then a three permutation must be chosen either from the multi-set or from the multi-set, which gives us, if we have two a's, then our three permutations must be chosen from, giving us, and the sum will give us the number of three permutations of this multi-set. And this suggests a formula for computing permutations of multi-sets. The number of k permutations of the multi-set is the sum of all possible expressions of this form, where the individual factorials are less than or equal to the individual repetition numbers, and k is the sum of the p a's. And again, it's important to remember, don't memorize formulas, understand concepts. And especially in this case, this is a formula that is almost never practical to use. Instead, what's going to be important is this idea of summing over all possible expressions of a particular form. To see why this formula is actually kind of useless, let's consider a fairly modest problem, finding the number of five permutations of this multi-set. Since there are four elements, we want to list all possible p one through p four, where the p's are less than the corresponding repetition numbers, and they add to five. Now, to keep track of everything, we can write these as ordered four tuples. And to be systematic, let's start with the fewest a, b, and c possible. If we have zero a's, zero b's, and zero c's, then, since this is supposed to be a five permutation, we have to have five d's. But we can't, since our set only has two d's. And in fact, since we only have two d's, we need at least three c's. So we can have a zero a, zero b, three c, and two d, which we'll record as the ordered tuple. And this will give us permutations. We could also have four c's and a d giving us a tuple, which gives us... We can't have five c's, so now let's increase the number of b's. If we have zero a, one b, then we need four more from c or d. And we can do this with two c's and two d's, three c's and a d, or four c's, which give us permutations. If we increase the number of b's, so we now have zero a's and two b's, we need three from c or d. And we can do this using, which give us an additional... We can't have three b's, so let's start with one a, zero b's, and four from c or d, which we can do as giving us the possibilities for one a, one b, and three from c or d are with permutations. The possibilities for one a, two b, and two more from c or d will be with permutations. Two a, zero b, and three from c or d give us two a, one b, and two from c or d are. Two a, two b, we could have three a, and two from c or d. Three a, one b, and finally we could have three a, and two b with none from c or d. This rather horrifying expression is what we end up with when we try to use the formula. And so the natural question to ask is, is there an easier way? And the answer is, we'll take a look at that next.