 All right, so we've seen that permutations are a useful way of calculating how many different arrangements there are of indistinguishable objects, but when we're thinking about chemistry problems, I'm sorry, of distinguishable objects. Permutations help us calculate the number of arrangements of distinguishable objects. When it comes to chemistry problems, however, it's much more common for objects to be indistinguishable from one another. I can't tell this electron apart from that one, or this hydrogen atom apart from that one, or this benzene molecule apart from that one. So objects are more commonly, or at least very commonly indistinguishable from one another, and we need to be able to calculate the number of ways of arranging those objects when they're indistinguishable. So, as the example we'll use in this case, starting with an easy question, relatively easy that doesn't involve any complicated chemistry terminology, the indistinguishable objects that we'll talk about are these filled circles. So let's say I've got two of these circles, and I'd like to place those circles into a collection of five boxes. So I have five boxes. I want to put those two circles into those five boxes anywhere I wish. Each box can only fit one circle, but I have choices about where to put them, and I would like to know how many different ways are there, what's the multiplicity of arrangements of those two circles in five boxes? So this problem is simple enough, we can write down all the possibilities. So let's do that relatively quickly. So I've got five boxes. I can fill the first and the second box with two circles, and then I've run out of circles, and that's one of my options. Or perhaps after putting the first one in the first box, I decide to put the next one in the third box, or I could have put it into the fourth box, or I could have put the next one into the fifth box. So that's certainly four different options. I also could have not put the first one into the first box at all, but into the second box, and then put the next one in the third, or the fourth, or the fifth. So you'll notice I didn't write the option of putting the first one in the second box, the second one in the first box, because that would have given me the same result as one that I had already written down, because these two circles are indistinguishable from one another. It doesn't matter what I put one here and then one here, or the other way around, they get the same result. But I do still have more options to consider. I can put one into the third box and one into the fourth. I can put one into the third box and one into the fifth, and I can put one into the fourth, one into the fifth. So I've tried to be somewhat systematic about the way I arranged them and the order in which I discussed them to make sure I haven't forgotten any possibilities. But we can sit here and think about it more and write down more possibilities, but these are the only possibilities that exist for where I can put these two circles into five boxes. So if I count those up, one, two, three, four, five, six, seven, eight, nine, ten, I find ten different choices for how to put two objects into five, two indistinguishable objects into five boxes, a multiplicity of ten. So that's fine for problems that are small enough to do by hand. Got a little bit tedious even for this small problem. But we'd like to know how to do that for a more complicated case where I've got more indistinguishable objects. Let's say a mole of benzene molecules that I'm putting into various locations in some chemistry problems. So we clearly can't write that problem out by hand. So we're gonna need to know how to do this for much larger numbers. So let's think about where we got this answer a little more systematically. And I can say one way of getting that answer was to say, pretty close to what I did here is I had five choices for where to put the first circle. I could have put in box number one, two, three, four, or five. And then once I had put it down, I had used up one of the boxes. There's only four boxes left. So there's only four options left for where to put the next circle. So it's beginning to look like a factorial. But notice that I stopped there. I only put two circles into the boxes. So five choices for the first one, four choices for the next one. I don't continue because there's no more circles. So it's just five times four. That would give me 20, 20 different ways of arranging the circles in the boxes. That's already bigger than the answer we found. So what went wrong? Why am I not getting the same answer here as I did here? It's because the objects are indistinguishable. As we mentioned before, it doesn't matter whether I first put one in box number one and then box number two, or I first put in box number two and then box number one, I get the same result either way. Because the two circles are indistinguishable, there's two different ways of arranging the circles. So I need to divide by two to fix for the fact that I've over counted by multiplying five times four. I've counted this and this as different arrangements, even though they're indistinguishable from one another. So I need to divide by two to fix that and I get 10. So that gives me the right answer. Now we're on the right track to finding out what it is we're supposed to do to calculate the multiplicity for the more general case. Before I write down a more general formula, let me first write down a different example. So let's say we're done with this example and I give you another example. Let's think again about rearranging letters. Let's say I have five letters, two of which are the letter A, three of which are the letter B, and I want to know how many different arrangements of those letters there are. I could sit down and we can write out all the different possibilities, but we can save ourselves that work by noticing that this is actually exactly the same problem as the circles and boxes problem. I have five objects, only two of which are A's. The real question is in which letters, first and second, first and third, first and fourth, in which letters am I placing the A's? Placing the A's in the five letter word is exactly the same as placing the circles in the five boxes. So this is also gonna give me W equals 10 because it's the same problem. It feels a little bit like a different problem. There's two very distinct types of problems that give you the same answer that have slightly different feeling to them. One is this type of problem if I have n indistinguishable objects. Let's go ahead and say indistinguishable. And I want to distribute those across n locations. In a chemistry context that might be like I have little n molecules and I have big n places on a surface on which to put them or I have certain number of electrons that can fill certain number of energy levels in a molecule. A different type of problem that's really the same is this problem. Let's say I have n total objects like these letters A's and B's. Some amount of them, little n of them over one type. I've got two that are letter A's and the rest of them and minus n of them of some different type. And when I say type, the ones that are of type A, those are indistinguishable from one another. The ones that look like a B, those are indistinguishable from another. So this is like a shuffling problem. How many different ways do I shuffle these objects around? In a chemistry context that might be something like if I have certain number of molecules that are cis and a certain number of molecules that are trans, how many different ways can I arrange them? But it's important to recognize that those two flavors of problem are exactly the same. So now we're ready to write down the equation that we would use to calculate the multiplicity for either this type of problem or this type of problem. So generalizing from five boxes and two circles or five letters with two of them of the first type. If I in general have n letters, if they were all different, then I'd have n factorial ways of shuffling them around. For a shuffling problem, I'd have n factorial ways if they were distinguishable, right? But they're not distinguishable. Once I've written down one of these ways of writing down these five objects, the two A's are indistinguishable from one another. So there's two factorial ways of shuffling those A's around within themselves that doesn't change what the final answer looks like. And also, the B's are indistinguishable. There's big n minus little n of these, so n total objects, little n of them are A's, the rest of them, n minus n are B's. And there's n minus little n factorial ways of shuffling the B's around amongst themselves that I can't tell the difference between because the B's are indistinguishable. So this quantity, big n factorial over little n factorial, and then divided again by big n minus little n quantity factorial. That will be the number of ways of shuffling the objects around. This quantity has a name. It's called a binomial coefficient. And let's see, just to make sure that we're getting the right answer for problems like this, when we use the binomial coefficient, we can go back to our two circles in five boxes problem and say, what would we get if we use this expression? The multiplicity here, big n, the number of boxes is five. So five factorial in the numerator. I divide that by a two factorial for the number of indistinguishable circles there are. I divide it also by n minus little n, or three factorial. That accounts in this type of problem and accounts for the fact that the B's are indistinguishable. That's like saying either the empty boxes are indistinguishable or it's like saying once I place two circles and the other three boxes were empty, I don't continue with a factorial in the numerator anymore. In other words, when I write out this factorial in the numerator, five times four times three times two times one in the denominator. The three factorial in the denominator, the three two one down there, cancels exactly the three two one which is part of the denominator upstairs. So that's why, from one point of view, that's why we only did five times four in the numerator. So the five times four in the numerator is only part of the factorial. And then we divide by two to fix the over counting of the circles. But either way, that still comes out to the same answer, which is ten. So now, we have an expression that we can use to calculate the multiplicity. So let's do one real chemistry example with this equation. So now let's say we have a polymer. Let's say I have a polymer and there's some double bonds in this polymer. And every time I have a double bond, I can either have a trans linkage or I can have a cis linkage in this polymer. So let's say I've got eight different linkages. So it's not a very long polymer. It's only an oligomer with nine units and eight linkages between them. So I've got eight places where this polymer can either be cis or can be trans. So let's go ahead and write down cis, trans. Let's say I wanna consider the possibility that four of those eight linkages are cis. How many different ways are there that my polymer with eight linkages can have four cis linkages and four trans linkages? So what's the multiplicity? That would be eight factorial over little n factorial. And then big n minus little n is also four. So I need to know what is eight factorial over four factorial, four factorial. Let me, I suppose, take this opportunity to point out that when we get tired of writing down factorials over factorials, there's one bit of shorthand notation that we often use. It's common enough that we use these binomial coefficients to say big n factorial over two smaller numbers in the denominator that add up to big n, that the shorthand way of writing those two down is writing big n above a little n with no line between them. This is not a fraction and parentheses around them. So that's also a way of writing the exact same quantity called the binomial coefficient and we call that with the way we describe that in words is we say n choose n because what we're actually trying to do is saying out of five boxes how many of the ways are there of choosing where two indistinguishable objects go or out of eight linkages how many ways are there of choosing which four to make cis linkages. So I could have written this down as eight choose four. But the way I actually evaluate it is eight factorial over four factorial, four factorial. Numerically that's eight times seven times six times five and then four three two one is going to cancel the four three two one from one of these four factorials. The other four factorial I'll write out as four times three times two times one. Numerically the four and the two cancel the eight. The three turns this six into a two. So I've got two times five is ten times a seven. That gives me 70. So if all I wanted to know was how many different polymers can I make with eight linkages where four of them are cis there's 70 different of those polymers or ligaments that I can make. The last thing I'll point out is that typically the reason we calculate multiplicity is not really because we're interested in how many ways I can make the polymer but if I just randomly make a polymer and I'm interested in what is the likelihood that it ended up with four cis out of eight linkages then the multiplicity lets me calculate that sort of thing. So as a second part of this problem the probability that I were to get four cis linkages out of my eight would be assuming that every possibility is equally likely I can calculate the multiplicity. How many ways are there of getting four divided by the total multiplicity? How many total ways of connecting the polymer do I have? We've just calculated the number of ways of getting four cis linkages would be 70 the total number of polymers I can make that's back to a different type of problem linkage number one could either be cis or trans two options linkage number two could also be cis or trans so there's two options for the first linkage two options for the second linkage two options for the third linkage all together for the eight linkages there's two times two times two to the eighth or 256 total ways to build that polymer so 70 out of 256 of those ways to build that polymer I'll go ahead and put a number on that that's close to 27% so that's beginning to sound like a chemically interesting number to calculate if I have an equal chance of making a trans or a cis linkage what's the probability that if I make an oligomer with eight linkages that four of them half of them end up cis it's about 27% so this begins to show us how we can use this idea of combinations to calculate either multiplicities or to calculate probabilities of events