 Welcome to our lecture series, Math 3120, Transition to Advanced Mathematics for Student Sets at the University. As usual, be your professor today, Dr. Andrew Misseldine. Previously in our lecture series, you've talked a lot about sets, talked a ton about sets, but we've also talked about the related notion of a list, which remember the main difference between a set and a list was it had to do with this idea of order, right? In a set, the elements could be arranged in any order whatsoever, and that doesn't give you a different set. As opposed to a list, order was paramount. That was the defining characteristic of a list compared to a set. We often refer to them as ordered list, but honestly, the word ordered is redundant in our vocabulary because order is inherent to the notion of a list. There is another quality that does distinguish between a set and a list, potentially the idea of repetition. When we discussed sets earlier in our lecture series, we said that if an element shows up more than once in the set, we ignore the multiplicity, the repetition is relevant. But for a list, the repetition is important because if the same element shows up more than once, but since the list is ordered, it could show up in the first spot, it could show up in the third spot, we have to distinguish between those cases. Right now, we want to introduce a little bit of vocabulary as we consider it ordered list. What if we care for repetition? What if we don't? Because we've seen some commentatorial problems where we allow for repetition in our list, and sometimes we don't. So some vocabulary could help us with that. We say an ordered list with repetition is called a string. This is just borrowed from computer programming, for which a string is going to then be this. It's an ordered list of characters of some kind. And we do allow repetition in that situation. If we don't allow for repetition, our list instead we refer to as a permutation. To be technical, there are other meanings of permutations in mathematics among commentators itself. But at this stage of the game, people often use the term permutation to describe this ordered list without repetition. If ever we had some confusion with the future notion of a permutation, which is a bijective function from a set back into itself, we might actually use the word partial permutation because the permutation doesn't necessarily include every element of the alphabet we're drawing from, but typically we'll refer to this just as a permutation. And I want to confess that as we've looked at multiplicative counting principle like problems before, that's exactly what we've been doing. We've been trying to count strings and we've been trying to count permutations. So let's see some examples of these. These won't feel very different. Let's say in the state of Georgia, license plates consist of four digits followed by a space and then by three capital letters. So you might get something like one, two, three, four, four digits and a break, ABC. So this would be like a typical Georgian license plate. If the first digit cannot be zero, how many different license plates could you possibly have here? So much like our example, we have four buckets that have to be filled with digits and then we have three buckets that have to be filled with a Roman character of some kind. Now for these buckets and the first position, because we're looking for strings in this situation, repetition is allowed, it doesn't say to the contrary. And while there are different alphabets for the first four letters and the last three characters there, this is still an example of a string, a little bit of restriction on the alphabets but it still is a string. It's an ordered list repetition is allowed. Now, because zero can't show up in the first digit, we only have nine options for the first one, okay? But because we have repetition now, we could reuse the first digit in the second digit, but now zero is an option. So you get 10 options for each of the subsequent digits there. And then for the Roman letters that show up at the end, there's 26 there as well. And so the number of possibilities you're gonna get is nine times 10 cubed times 26 cubed as the total number of license plates, which if you were to multiply those things out, you're gonna get 158,184,000 license plates in the state of Georgia there. Now, I want you to be aware that when you work with combinatorial problems, oftentimes the expanded answer is not of any interest to us whatsoever. We are much more interested in this right here. Where did this number come from? Because anyone with a calculator can get from here to here, that's no problem at all. But understanding that this counts the number of strings that satisfy the conditions above is the essence of what our combinatorial exercises are about. Now, I want to revisit this same problem but make one slight tweak. Suppose we're looking at Georgian license plates, but what if repeated letters are not allowed? If repetition is not allowed this time, how does that change the problem? It doesn't really change much, but instead of counting strings, we're now trying to count permutations. We still have four buckets that have to be filled for the digits, and we have three buckets that have to be filled with the Roman letters. The Roman letters are gonna be easy to do, which is why I'm gonna start with them. For the first one, we have 26 letters that were allowed, but without repetition, you get 25 and 24. We've seen examples like this before. Coming back to the Georgian license plate, the first digit can be anything except for zero, so you have nine options there. Then when we come to the second one here, we no longer have repetition. So whatever letter we use in the first spot, we can't use it for the second spot. So we normally would think we go down, but zero is also included, so we lose one option from the first spot, but we gain one option because we have now zero. So in this case, nine is also the options for the second one. They're not the same nine options, but there are nine options for the second digit. But then whatever one we used here and used here can't be used again, so you get eight and seven, like we've counted these things before. And so then the total answer would be nine times nine, times eight, times seven, times 26, times 25, times 24, which of course we can multiply that out and we'll get 70,761,600. But again, it's not the expanded answer that matters. This is the one we're gonna care about right here. But when you look at an answer like that, when you count permutations, because the numbers are often decreasing by one, it's not as easy to write in this more compact exponential form. And so a little bit of notation can help us save the day, so we don't have to write these expanded things all the time. So let's remind ourselves that when we have any pods of integer n, we can define the in factorial, which is denoted as in with an exclamation point, some people call it a bang, in bang, in factorial here. This is the product of all consecutive integers from one up to n. So you get n times n minus one, times n minus two, times n minus three, all the way down to three times two times one, okay? We do want the in factorial function to be defined for all natural numbers. So we also do define zero factorial, and that's defined to equal one. We'll give you some explanation about what this is and why we define zero factorial in one in just a second. But this is a notation you've probably seen before. Generalizing the notion of in factorial, we wanna introduce a symbol which we're gonna call the falling factorial, okay? The falling factorial, which in our lecture series, we're gonna denote this where if k is a natural number, in addition to n being a natural number, n could be zero here as well, k could be zero. We then will define the symbol n, and then the n is gonna be inside of parentheses and we're gonna have a subscript of a k, for which this will be our falling factorial. What this number means is you're gonna take the product where you start with n, then you times that by n minus one, you times that by n minus two, and you're gonna times that all the way by n minus k plus one, which using notation we've seen before. I want you to note here, we can write this as a product, a generalized product where you're gonna take n minus i here, where the i will range from zero up to k minus one. This is the same thing as the notation we've defined right here. You can also define this symbol, the falling factorial using factorials itself, where you take n factorial and you divide it by n minus k factorial. Now, I wanna mention that this notation for the falling factorial is not universal. Other notations that exist, a very popular one as you write things like npk, many scientific calculators, this is the notation they use, where you have a subscript of n that precedes the p and then there's a subscript of k that follows it. The capital p here stands for permutation because falling factorials, like we saw in the previous example, falling factorials are used to compute permutations that is list without repetition, because the answer to the previous problem we could have written it as nine times, we would then take nine falling factorial by three and then you times that by 26 falling factorial by three as well. We can express answers using these falling factorial. So this npk button that you might see on the calculator, this is an alternative notation. Some people write this as p in comma k. And honestly, because in mathematical type setting, putting a subscript before a character can be really tedious to type up. So some people use this notation as well. This, again, there's no universal notation, but something like this is far less used in comic torques. This is a more standard notation for more advanced mathematics. Now, if we were to come to a problem like the following, and how many ways can five people be lined up? I wanna make mention that if you're in kindergarten, this is a fundamentally important question, right? Because when you're in kindergarten, you wanna be the line leader. The line leader is the first kid to go to the bathroom, the first kid to get a drink, the first kid to go to lunch for the day. So the order in which the kids are lined up is super, super important. Suppose we have a class with only five students in it, right? How many people can be the line leader? Well, there's five options for that, okay? But then for this, the next position might as also pretty important because the person behind the line leader will be the line leader for tomorrow. So they're pretty much caring about that. You have four options for that. The status then of course goes down the line. And what you see here is that you have to, if you have to arrange five people into a line where the order matters because we care who's line leader, we care who's gonna be line leader tomorrow and who's gonna be the line leader after that. The status matters based upon how you order them. If you align all things in your set, then you get five times four times three times two times one. There's gonna be five factorial ways of lining up five people into a line, which of course is the same thing as 120 ways. But again, five factorial is what we care about. So when you are counting a list, all the possible list without repetition and you use every element inside of your alphabet, then you end up with a factorial itself. And this gives us an example of a permutation. Now, if you don't use up all of the elements in your set, like with our Georgian license plates from before, you do get a falling factor. You'll go all the way down to one, but you do get this product of decreasing numbers. Like with our Georgian license plates, you only have three Roman letters. So if you don't allow repetition, you still can't use up the whole alphabet, there's 26. And this is why we call it a partial permutation. You don't use all of it, but like I said, the adjective partial is generally dropped and we just call it a permutation. Let's look at one more example of these permutation type problems, which case, let's suppose that it's time to elect four class officers for our combinatorics class per se. And pretend this class has 80 students in it and we have to elect a president, a vice president, a secretary and a treasurer. Apparently in combinatorics, we take things super seriously that we even have to have officers in the class. But hey, with 80 students, that's a lot of combinatoric students. How many ways can we choose these? Well, I want you to be aware that if we decide how many different officers can we select here, the order in which we choose the matters, because being president is not the same thing as being vice president, which is not the same thing as secretary or treasurer as well. And so as we then decide how many ways can we choose the different officers, well, you are going to get 80 options for the president and then what comes next, you're going to get 79 options for vice president because whoever we choose for president can't be vice president. Repetition is not allowed here because only one person can hold each office. And then that would give you 78 for secretary and 77 for the treasurer right there. And so the number of ways we could select these four officers from the 80 students is then going to turn out to be 80 factorial fall by four like so, so 80 fall four is the number which of course that's 80 times 79 times 78 times 77 and if we want this in expanded form this would be 37,950,920 possibilities. Now with previous Covenant Torx problems we saw we might be tempted to list all the possibilities but you can see that with even small problems like this this number just gets too big, the list gets too long we can't consider every possibility but the Covenant Torx techniques of permutations and factorials can help us and particularly as falling factorial is a very useful tool when you have a list without repetition.