 In a set, repeated elements are not permitted. So we could have a set of friends, since each person is distinct, this is a set. Or we could have a set of silverware, one fork, knife, and spoon is a set. But a set of four forks is not a set, instead it is a multi-set. In a multi-set, we could list the repeated elements. So a set with four a's, two b's, and three c's would be, and as with sets, the order we list elements doesn't matter, so this could also be. However, we usually indicate the number of each type. So we have four a, two b, and three c, where four, two, and three are referred to as the repetition numbers. If we want to leave the number of each element unspecified, we'll use numbers as the elements. So here we have n-one-ones, n-two-twos, n-three-threes, and so on. And this helps avoid confusing expressions like n-one-a's, n-two-b's, and n-i-z's. If we know both the number of each element and the number of types, we'll continue to write things like five a's, two b's, and three c's, unless we forget. So we could ask, given a multi-set, how many combinations can we make of a given size? But this is too hard. To start with, we'll get there eventually. Instead we'll ask, given a multi-set, how many permutations can we make of a given size? Suppose we have a multi-set and want to form a permutation of n elements from the multi-set. We'll call this an n permutation. There are two possibilities. We have enough of every type of object so we won't run out, or we might run out of some objects. Suppose we want to find the number of n permutations on the multi-set, with i distinct elements, one through i. If nk is greater than or equal to n for all k, then we won't run out of any element. And so by the multiplication principle, we have our first, second, all the way up to our nth choice, and we always have any one of the i different elements. So we have i possibilities for the first choice, i for the second, and so on up to the nth choice, and so we have i to power n permutations. So how many i permutations can you form using any number of a's, b's, and c's, but at most five d's? Note that we can describe this as the five permutations from the multi-set, an infinite number of a's, an infinite number of b's, an infinite number of c's, but only five d's. We won't run out of a's, b's, or c's, and since we'll need at most five d's, we won't run out of d's either. And since we won't run out of any element, the number of five permutations will be, well there's five choices, and four possibilities for each choice. So we get four to the fifth permutations. How about nine permutations of this multi-set? Since nine is greater than four, two, and three, we could run out of any of the elements. And since four plus two plus three is nine, we in fact must use all of each type of element. Now in mathematics it's sometimes useful to engage in wishful thinking. If all the elements were different, there would be nine factorial permutations. Let's see if we could transition from the multi-set to a set where all the elements are different. So let's consider a permutation of the multi-set, for example. If the a's were actually different, say this one permutation of the multi-set would correspond to four factorial permutations of the multi-set with different a's. Like, so if we multiply the number of permutations of the multi-set by four factorial, we'll get the number of permutations of the multi-set where the a's are distinct. But wait, there's more. Now consider any permutation of this multi-set. If the b's were different, say, then any permutation of this multi-set actually corresponds to two permutations of a multi-set where the b's are distinct. So if we multiply the number of permutations by two factorial, we get the number of permutations in the multi-set where the b's are distinct. And again, any permutation of the multi-set would produce three factorial permutations of the multi-set where the c's are distinct. So if we multiply our permutations by three factorial, we'll get permutations of our new set. But notice that our elements are distinct. And since the elements are distinct, the number of permutations of our final set is nine factorial and so we can find the number of permutations of our original set as... And this suggests the following. Given a multi-set where n is the sum of the repetition numbers, then the number of n permutations of m times the product of the factorials of the repetition numbers will equal n factorial. And we could turn this into a formula, but remember, understand concepts. Don't memorize formulas. And the important concept here is we took our multi-set and gradually transformed it into a set whose permutations could be computed easily. Ah, well, here's the formula anyway. So a polypeptide chain is a sequence of amino acids. How many different polypeptide chains can be formed from three arginines, five lysines, and six glutamines? So here we're forming a polypeptide chain for the multi-set 3a's, 5l's, 6g's. And since we're using all of them, we're actually forming a 14 permutation. And a theorem says the number of 14 permutations will be 14 factorial divided by 3 factorial, 5 factorial, 6 factorial.