 So now we're ready to start considering a problem that turns out to be very important for describing the number of ways molecules can arrange themselves in a chemical system. And that's this question of permutations. So to explain this idea, we'll start with an example. Let's say we have three letters, A, B, and C. And I want to know how many different words, not English words, but how many different arrangements of those three letters, how many different three-letter combinations of A, B, and C I can make. If I just have three letters and I can move them around, I don't get to make more copies of any of them. So how many different ways are there of shuffling those letters around? So for a small problem like this with only three letters, we can do that by simply writing down all the possibilities. We call that doing it by direct enumeration. If I directly list all the different possibilities, certainly I can list them in the same order I was given them, A, B, and C. Or I can keep the A at the front and swap the B and the C. So that would give me A, C, B. And that's the only ways of doing it with an A in the front. But I can put another letter at the front. I can put a B at the front and then write the other two letters. Or I can swap the other two letters. Or instead of putting the B at the front, I could have put the A, I'm sorry, the C at the front with the other two letters in either of the two different orders. And no matter how hard I think about it, I'm not going to be able to come up with any other way of writing these three letters in any different order. There's only, so if I look at those, there's only six different ways of combining those letters in different orderings. So we say that there's six different permutations of these three letters, A, B, and C. So for small problems, it's simple enough to write down all the possibilities, but it doesn't take a very large problem before that becomes either tedious or completely impractical. So if you want to convince yourself that that's true, if you just think about the problem with six letters, say A, B, C, D, E, and F. If that sounds like it's perfectly doable by this direct enumeration approach, I suggest you go ahead and pause the video. Write down all the permutations that you can think of for the six letters A, B, C, D, E, and F. And then when you get bored of doing that, come back and we'll talk about a more efficient way of doing it. So whether you didn't go away or whether you paused it and tried it and came back, we need to know whether we need to know a way of calculating these number of permutations without writing all the possibilities down. So to do that, we'll think about the problem this way. We have three letters. They have to go in three spots. First letter, second letter, third letter. When I started writing down possibilities, there were only three choices since I have three letters. There's only three choices A or B or C for what I could have put in the first slot. So once I had put a letter there, maybe an A or maybe a B or maybe a C, whichever letter I decided to put there, that used that letter up. There's only two choices left. So what letter I chose to put in the second slot was restricted to only one of the two that I had left. So maybe that was a C and then once I've written those two down, then there's only one choice left for what to put in the last slot. So A was my only choice. So regardless of what numbers I chose to write down there, there were three choices I had to make in the first slot, two in the second slot, only one in the third slot. So the reason I have a total of six different permutations is three choices to make and then for any one of those, I could have made two other choices and then for any one of those, there was only one choice left in the last slot. So that's another way of obtaining that number six. More generally, if I have N objects that I want to find the permutations of, the six letters A, B, C, D, E, F, for example, I'm not patient enough to sit down there and write out all the possibilities. There's going to be six choices. Let's not do it for six. If I did it with six letters, there'd be six choices for the first letter, then five, four, three, two, one. So I just keep decreasing until I run out of places to put letters. But in general, if I have N objects, there's N a choice of any one of N of them that I could have put for the first letter and I've only got N minus one of them left for the second letter, N minus two of them left after that and keep going until I get down to one. So the product of all the integers from one to three all the way up to the number of objects that I have, that's how we calculate the number of ways of arranging those objects, the number of permutations. A shorter way of writing that down is to write this formula in this form using what's called product notation. So that deserves a brief comment. You're probably familiar with summation notation. If I were to write the sum as K runs from one to N of something, some terms X sub K, what that means is X1 added to X2 all the way up to XN. I sum up all the terms X1, X2, X3, X4 all the way up to N. So that's what this big sigma for S means is summation notation. Product notation is the same idea except instead of if I put the sign out front as a capital pi, standing for product notation, then what that means is I take the product of X1 times X2 times, etc., all the way up to times XN. Let's use a dot instead of an X. So multiply X1, X2, X3 all the way up to XN, and that's what I mean when I use this product notation. So you've likely seen summation notation before, whether you're comfortable with it or not is a different question, but product notation you may or may not have seen before just means take the product of all these things. So a much shorter way of writing 1 times 2 times 3 times all the way up to N without being forced to use this dot dot dot is to say take the product of all the numbers from 1 running up through N. And because even that is a little cumbersome sometimes, we can also write down the same thing this way. So instead of writing the product of all the numbers from 1 through N, I can write it instead as N with an exclamation point where this is called factorial notation. And it means exactly the same thing. If I say calculate N factorial, that just means calculate the product of 1 times 2 times 3 running all the way up until N at the top end and that's called calculating the factorial. So whichever way you want to think about it, an explicit product written out using product notation or written with this factorial notation, the number of ways of permuting N objects is N factorial or the product of 1 up through N. And what we're talking about terminology, this quantity N that we're talking about, the number of ways of doing something, in this case the number of ways of permuting these letters or these objects, the fancy word for the number of ways we can do something is the multiplicity, the total number of different ways of arranging something or the total different ways that something can exist. So we have this method of calculating factorials. So as a last example, now that we've introduced the terminology, we can do a problem that relates to chemistry a little bit. So let's say we're talking not about letters and words, but let's say we're talking about DNA. So I have, DNA has base pairs. Let's say I have exactly one adenine, one guanine, one thymine, one cytosine. I have exactly one each of each of these bases. And I want to make a single strand, a tetramer of single stranded DNA. I want to connect these four bases together in some order. So the question is how many, what's the multiplicity? How many different tetramers of single stranded DNA can I form out of these four individual base pairs? So that question has a lot more chemistry terminology in it. But it's really just the same thing as asking how many different words can I spell with these four letters? The total multiplicity of arranging, of permuting these four bases in any order that I want is going to be four factorial or four times three times two times one, if I want to write out the product. So four times three is 12, times two is 24. So there's 24 different permutations of these four DNA bases. So that just gives an example of a chemistry problem. We do have one important issue left to discuss, which is everything I've written down so far. This evaluation that the multiplicity, the number of ways of permuting objects is n factorial. That is all true only if the letters or the base pairs or whatever objects it is that I'm arranging, if they're different from one another. Or as a terminology word, if they're distinguishable from one another. For example, we won't solve this example, but if I pose the question, suppose the letters weren't all different. Suppose my three letter word wasn't A, B, C, but suppose it was A, A, B. If I ask how many different ways are there of arranging those letters, A, A, B, or two of the letters are the same? The answer is not three factorial anymore. There's not six different ways of arranging these letters. I guess we will go ahead and solve the problem. I can write A, A, B, or I can write A, B, A, or I can write B, A, A. Those are the only choices. If I try to write anything else down, I've written down one of the ones that I've already written. There's only three different ways of arranging those objects, and I don't get the same answer of six because the two As are identical to one another. What we've learned so far only works for distinguishable objects. If I have n distinguishable objects and objects that can be told apart from one another, then the number of permutations of those objects is n factorial. The next video lecture will cover how to deal with problems where some of those objects are not distinguishable from one another.