 So we've seen that the binomial coefficient and the binomial probability distribution can be useful for calculating the number of ways something can happen or the probability that will happen if there's two different outcomes. And sometimes that's useful if we have two outcomes like spin up or spin down for an electron or cis or trans as the only two conformational states we're interested in. But oftentimes there's more than two options to consider. So we need to consider not just the binomial coefficient or binomial probability distribution but the multinomial coefficient. So to give you an idea of a problem where that might be relevant, let's do an example, a DNA example. So let's say we have 10 different base pairs of DNA, not base pairs, 10 different bases. Let's say we have 4a's and 3 anines, 3 guanines, 2 thymines, and one cytosine. So the total of 10 different bases, 4 plus 3 plus 2 plus 1 is 10. And I want to know how many different ways are there, how many different 10 mers of DNA can I assemble out of these 10 individual monomers. So I could put them together in this order, some other order, how many different combinations are there. If they are distinguishable, if every individual monomer was different from all the others, then I could arrange them in 10 factorial different ways. That's what our permutation is for, is calculating the number of arrangements of distinguishable objects. But these adenines are indistinguishable from one another, the 3 guanines are indistinguishable from one another. So there's a lot of ways of writing down A's and G's and T's and C's that are the same as each other because I can mix and match the A's and I can't tell that I've swapped them. So 10 factorial is not the right answer. If I write down any particular, let's go ahead and write down one orientation, A, A, G, C, T, A, G, G, and then I need a T. So there's one of the orientations, I've only written down three A's, so let me write down a fourth A. There's one of my orientations of these 10 bases. If I switch these two A's, I get the exact same result, or if I switch these two A's. There are different A's that I can shuffle around in four factorial different ways, so I have to correct by a factor of four factorial because I have four indistinguishable A's to shuffle around. I also have one, two, three different G's that are indistinguishable that I can shuffle around in three factorial ways. And I have two thymines that I can shuffle around and one cytosine that I can shuffle around. So these factorials in the denominator are correcting for the fact that the four A's are indistinguishable, the three G's are indistinguishable, the two T's are indistinguishable, and then the one C is indistinguishable from itself. It doesn't matter whether you do or don't include a one factorial in the denominator. So numerically, we can go ahead and figure out what that number is, 10 factorial in the numerator. I divide that by 4, 3, 2, 1 for the A's, 3, 2, 1 for the G's, 2 and a 1 for the T's and a 1 for the A's, so the 1's don't really matter. The 4, 3, 2, 1 kills the term up there. If we were to do it by hand, the 6's kill 3 and 2 kill 6 and the 2 knocks this down, let's say, to a 4. If I multiply those out, 10 times 9 times 4 times 7 times 5, that will come out to be 12,600. So for the numerical answer to this particular question, there's 12,600 ways of shuffling 4 A's, 3 G's, 2 T's and a C around, so there's 12,600 distinct 10 MERS of these bases of DNA. More generally, this is an example of a question where I say I have a total of N objects, 4 of this type, 3 of this type, 2 of this type and so on, so in general I might have N sub 1 of type 1 that are all indistinguishable from one another. I can tell 1's apart from 2's, but I have N2 of type 2 that are indistinguishable from one another and so on. And I would like to know what is the total number of ways of arranging those objects where some of them are indistinguishable from each other and I have more than just two types of object. This same process is going to work no matter how many different types of objects we have or how many of each type we have. So it would be N factorial if they were distinguishable, which they're not, so I have to divide by N1 factorial and 2 factorial and so on to correct for the fact that they're indistinguishable. So that's exactly what we did in this example and it would work in the general case. The shorthand notation we can use just like for the binomial coefficient, we have a shorthand and choose little n that we can use to write down a binomial coefficient. The shorthand we can use to write down this multinomial coefficient is big n on top, these little n's on bottom and 1 and 2 dot dot dot as many of them as we have. Put the little n's on the bottom for as many of them as types of objects that you have and that's our shorthand notation for the multinomial coefficient and I'll go ahead and label that the multinomial coefficient. If we don't like writing the dot dot dot another way we can write this is to say that this is n factorial making use of the product notation. What we have in the denominator is n1 factorial times n2 factorial all the way up to some final factorial for our last type of object. So this is the product of n something or other factorial, summing over all the different types of objects sub k. So if we want a more compact way to write this without the dot dot dot, we can use the product notation to do that. So this summarizes the multinomial coefficient. The next step will be using that multinomial coefficient to write down the multinomial probability distribution.