 So the branch of mathematics is called combinatorics, and as you might guess, this means that combinations are actually the more difficult thing to calculate. So in general, finding the number of permutations is easy. All we need to do is rely on the fundamental counting principle. But finding the number of combinations is so much more difficult that it gives its name to the entire field of study. The general approach we have to finding the number of computations is as following. In general, every combination that I am considering corresponds to several distinct permutations. So I can find the number of combinations in the following way. First I'll find the number of permutations, and then because each combination corresponds to a bunch of permutations, there's going to be some sort of over counting factor. And so I'll take the number of permutations and then divide by whatever this over counting factor is. So let's see how that might work. So for example, let's take our combination meal, and in this combination meal I can choose two of seven different sides. So how many different choice combinations of side dishes are there? And we'll make the assumption that the side dishes have to be different. Can't order double mashed potatoes, for example. Now let's go ahead and set this down. So I'll set down a table, and so I'll take my choices and my selections. And again, let's check to make sure that we know what we're dealing with here. So suppose I select French fries for my first side and salad as the second. So first off, are we actually dealing with a permutation or a combination? And so the question is, if I alter where I write the choice, does it change what I get? And well, it makes no difference. If I put salad in the first place and french fries in the second, it's the same as french fries in the first and salad in the second. It's the same set of side dishes. It makes no difference where I write the selections. So we're dealing with a combination. So our starting point, well, let's start with what's easier. So let's figure out the number of permutations that we have. So choices for our first side. Well, there's seven different sides I could choose. So for that first side, I have seven different possibilities. Now for the second side dish. Because the side dishes have to be different, there's our second assumption that side dishes must be different. So whatever I pick for my first side, there's only seven minus one. There's only six possibilities left for the second side dish. So I have seven choices for the first and only six for the second. Now this corresponds to the number of permutations that we have, where we would regard the choice as salad, french fries, as different from the choice french fries salad. So this product, seven times six, gives us the number of permutations. And so one way to approach this question is, well, suppose I choose these two as my side dishes. How many ways could I have chosen those two? Well, here's one. I could put the salad here, the french fries there, or I could put the french fries here and the salad there. Now rather than enumerating all those possibilities, which will take quite a bit of effort if there's more than two of them, I might approach the problem as follows. Now consider that first side. Well, having chosen the two sides, the one I actually write first, I have two choices for that first written side. So I have two choices of which side to write first, and then the second side that I wrote, well, I only have one possibility. It's whatever the other one is. So let's put that together. So if I think about choosing the two sides, there's seven times six. There's 42 permutations. However, that set of permutations will regard the choice salad french fries as different from the choice french fries salad. And in this case, there's two distinct permutations that correspond to the choice of side salad and french fries. So what does that mean? Well, those two permutations that correspond to one combination says that these 42 permutations are too many by a factor of two. So the actual number of combinations that I have is going to be the number of permutations divided by this over counting factor two times one equals two. And so I actually have 21 distinct combinations of two side dishes. Now, when we apply this to probability, it's convenient to look at this in a slightly different framework as follows, which is, if I imagine that what I'm choosing is not the objects themselves, but I'm choosing where they're going to be located. So for example, suppose I want to make a five letter word, which is putting five letters in some place, and I'm going to use the letters three a's and e and an m. So the question is, well, how many different words can I make this? So it's convenient to think about these choices that we're making is actually where we're going to place them, and we'll set down a table. So here's our letter, three a's and e and an m, and the number of choices I have for where to put them. And just to make this a little bit more clear, let's indicate where we're actually placing them. So for example, I might place the letters as follows. The first day I'm going to put in location three, the second a in location one, the third a location four, the first e location two, the m in location five. And if I do that, the word that I get, a-e-a-a-m. And if you can convince your opponent and scrabble that this is really a word, that will score you a couple of points. Now it's worth pointing out that in some cases I can actually switch where I write some of these numbers. So these numbers here, these three, one and four, it really doesn't matter where I put these three numbers. If I put this three over here and this one over there and move the four over, it's the same thing. So I might write this four over here and this three over there, I swap them, and I get the same thing. I still have an a in the first, third and fourth place, first, third and fourth places, an e in the second, an m in the fifth, and it's still the same word. So I haven't changed anything with that alteration of where I write the numbers. So what does that tell me? Well, I do have something of a combination here. However, if I switch these two numbers, so if I switch the two and the five, I end up with something different, a-m-a-e-e, and that's different from what I have here. And so here I can't switch these two without changing the result. I can switch among these three and still get the same thing. So here I have a combination, but there's also a permutation tied in with it. Most interesting problems have features of both combinations and permutations. So how do I answer the question? How many words can I make? Well, we can still approach it the same way. So how many choices do I have for where I put this? Well, there's five spaces in the word, so I have five choices there. I have four choices of where that a goes. I have three choices for that next letter. I have two choices for the next and one choice for that last letter. So now let's analyze this. So now I've picked places for these three a's. And one of the things that we determined earlier was that it really didn't matter what order I picked those places. So if I've picked position one, four, and two for the a's, well, it's the same as picking the positions four, one, and two, or two, four, and one, or a number of different possibilities. So how many ways could I have picked the three locations here? Well, there's three ways I could have picked that location for the first day. There's two ways for the second and one for the last one. So whatever positions I've picked here as a combination translate into three times two times one, six different permutations. How about that e? Well, I picked where it is here and there's no place else to put it. There's only one way I could have chosen that particular location. Likewise for the m, whenever I picked place or whatever place was left over, that's the one and only place that m could have been chosen. And so my permutation here corresponds to a combination here and the overcounting factor is this product, three, two, one, one, one. And so the number of words, the number of permutations, five, four, three, two, one, product divided by the overcounting factor three times two times one all over, and that works out to be 20 different words that I could make.