 Welcome back to our lecture series, Math 3120, Transition to Advanced Mathematics for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Missildine. We're gonna start lecture 14 with the following example, which will motivate what we're gonna do for the rest of this video. We want to list all combinations of the four objects A, B, C, D, such that in each combination, we're gonna take two of the elements at a time and repetition is not gonna be allowed here. Now, in Commentatorics, I want to mention the difference between a combination versus a permutation. When people talk about permutations, the order in which the element is listed matters. With combinations, the order doesn't matter how you list them. Now, in order to answer this question, what I'm first going to do is actually list all of the possible permutations of these things you can get. So for example, like what we've seen beforehand, if I take the falling factorial with four letters and two choices, this is the same thing as four times three, in which case there are 12 ways I can take with order two elements from this collection of four letters in the alphabet with no repetition. And you get something like the following. You would get A, B, B, A. You would get A, C, C, A. You would get A, D, D, A. You would get B, C, C, B. You would get B, D, D, B. And then finally, you would get C, D, and D, C. And that should be all 12 possibilities there. One, two, three, four, five, six, seven, eight, nine, 10, 11, 12. So we have all of them listed like so. Now you'll notice because of the underlines and perhaps just the way that I've listed all of these things, I put each and every one of them into pairs. Like notice I wrote A, B, and then B, A. Likewise, I wrote A, C, and then I wrote C, A like so. And then A, D, D, A, et cetera here. I actually wrote all of them in pairs so that if I care about the order, then these are the 12 lists we would get but the word combination doesn't care about order. So when it comes to a combination, you're only gonna get one for each of these pairs. You just get the pair A, B. You get the combo A, C, the combo A, D, the combo B, C, the combo B, D, and the combo C, D. Cause flipping the order does not change a combination. So if we don't care about the order, we only get half of the answers there. And in particular, the significance of cutting them half is actually it's a two factorial which so happens to likewise B, two. Because every ordered list, there are two factorial ways of writing them. So if you choose the letters A and B, you could write A first, then B second, or B first, then A second. So you get two times one ways of ordering them. And this one as well, and this one as well, et cetera. So in general, or maybe not in general yet, if you look at this problem, you take the total number of combinations which was the falling factorial fall, fall four choose two, in which case then by the division principle, you're gonna divide this by two factorial which cuts it in half because every thing got counted twice. Using the division principle, notice that we have this partition where things, if you ignore the order, go into the same cell. And so our group, our collection of list here, there are going to be the six cells in the partition. Where again, the partition groups everyone together if you don't care about the order. So A and B goes together, B and A goes together because if you switch, forget the order, they all go together. And each of these classes has a size of two. So if we take the set and then divide by the number in each of the cells, because each cell has the same size, this gives us the number of cells and that is the number of combinations here, okay? So I wanna make mention that with this example, if we had drawn the symbol like the following, we could have asked ourselves how many subsets of our set here do we have of size two? In which case each of these pairs, you can think of it as a subset. There's the subset that contains AB, the subset that contains AC, the subset that contains AD, the subset that contains BC, the subset that contains BD, and then the subset that contains CD. Because after all, that doesn't care about the order. And so when it comes to counting these so combinations, combinations are going to be unordered list without repetition. Counting combinations is the same thing as counting subsets, hence the title of this section of the textbook. And so generalizing the situation we saw in the previous example, we're now ready to discuss the so-called binomial coefficient. Suppose we arrange K objects chosen from a set of distinct of N, many distinct objects, if those arrangements are unordered without repetition. So that's the problem. What we're trying to do is we're trying to count combinations right now. That's what we say. And so we draw from a set of N objects and we're gonna choose K objects from that set of N objects. Now, clearly if K is bigger than N because you don't have any repetition, there's no way you could draw K things from N things without a repeat. So there would be zero ways of doing that one. So the case that we're really interested in is when K is a natural number, of course K and N are natural numbers in this consideration where K is a natural number and K is less than or equal to N. Then we introduce a symbol which this symbol is gonna look a lot like the multinomial coefficients we introduced in a previous lecture. But this time you're in, that's on the top, that's the same. But now you list only one number on the bottom, no separation by commas or anything like that. This is known as a binomial coefficient. The name of the objects are similar for a reason which we will talk about in just a second. By definition, the binomial coefficient will be the falling factorial N fall like K divided by K factorial. Which of course, as the falling factorial by definition is N factorial over N minus K factorial. When you put those things together, you get the formula, the binomial coefficient is N factorial over K factorial times N minus K factorial, okay? And so this binomial coefficient is counting the number of ways that you can choose K objects from a set of N objects. And like we mentioned on the previous slide, this question of counting combinations is equivalent to counting subsets. So if you have a finite set of cardinality N and you want to know how many subsets does this set have of cardinality K, the number of subsets will be exactly N choose K. That's how we read this number here, this binomial coefficient, we take N choose K. And the way we think about it is if there's N objects in the set to form a subset, we have to choose K of the objects to belong to the subset. Hence why we read this as N choose K. Now note, like I mentioned earlier, the binomial coefficients, they look like multinomial coefficients, at least the notation's similar. And the reason for that is that the binomial coefficients are actually special cases of the multinomial coefficients we saw earlier. Using the notation we had before, multinomial coefficients, you have the N on top, you have an N on the bottom, and excuse me, a K on the bottom, and then a N minus K on the bottom as well. Which of course by definition, this would be N factorial over K factorial and N minus K factorial. You'll notice that K plus N minus K is equal to N and you get the factors on the bottom. That's exactly what this is. This is a special case of the multinomial coefficients. The multinomial coefficients were based upon a permutation of course, right? A not permutation, a partition. That's the word I'm looking for. A partition, right? You have your cell and then you've subdivided into these smaller, you have your set and then you have subdivided into smaller cells, right? And then if you have N1 here, N2 here, N3 here, this is how we saw these things before. Now with the case of the binomial coefficient, we're taking one of the most basic of all partitions, okay? You have your set X and you have some subset A. Well, if you're not in A, then you're in the complement of A. So you're in one or the other. And so we talked about this with the subtraction principle because the additive principle of counting also, like the division principle, uses partitions to help you count a set. The subtraction principle came into play because every set has a very basic partition where you choose any non-empty proper subset, its complement together will form a partition. The binomial coefficients basically doing the same thing. If you take a two cell partition, the multinomial coefficient then simplifies to this binomial coefficient. And you're basically asking yourself, are you in or out? I remember there's this song from Aladdin III, the King of Thieves where one of the bad guys singing something like, are you in or out? I won't give you the whole song right here, but think about that when you think of binomial coefficients. You have to decide, are you in or are you out? When you build a subset, that's what you're doing. Are you in the subset or are you not in the subset? And so these multinomial coefficients are generalizations of these binomial coefficients, which is why of course they get their name. So then considering the following here, how many subsets of two elements can be chosen from a set of eight elements? The answer to that question right there is going to be eight choose two. Eight is the total number of elements in the set. Two is the cardinality of the subset. And so by definition, this is eight factorial over two factorial and six factorial. Six of course is eight minus two. In which case if we wanna compute this thing, notice that eight factorial is the same thing as eight times seven times six factorial over, well, two factorial is just a two and a six factorial. You notice on the top, I actually stopped at six factorial because as there's a six factorial on the top and bottom, they cancel out and this thing can simplify just to be eight times seven over two, which is four times seven. That is to say there are 28 such subsets. Subsets of cardinality two drawn from a set of cardinality eight. Similarly, how many subsets are there of cardinality three? That number is gonna be eight choose three. So this would be eight factorial over three factorial and five factorial. Always cancel out the bigger of the factorials on the bottom if you're computing this by hand. So you get eight times seven times six over three times two times one. And no matter what you do, this always turns out to be an integer. You can always find all of the advisors cancel out with something on the top. So you get eight times seven, which is gonna give you 56 subsets in that situation. There's 56, you can see why I'm not gonna list them all on the screen right here, but feel free to check it on your own. One more example of this, a set of size eight. If you want to count the subsets of size four, this would be eight choose four. Eight choose four looks like four factorial over four factorial. So we would look like eight times seven times six times five over four factorial, which is four times three times two times one, three and two is a six. Four goes into eight two times. So we're left with two times seven times five. Two times five is 10 times seven is 70. So we end up with 70 subsets in that situation. When it comes to factorial, these binomial coefficients, excuse me, one important thing to note here is that n choose k is actually the same thing as n choose n minus k. This is that whole in or out thing from before because if you choose someone to be in or you choose them to be out, it's basically the same choice because if I choose you to be, if I'm choosing k elements to be part of my subset, that means I'm choosing n minus k elements to not be part of my subset. And therefore, if you choose k elements to be part of a set, then you're choosing the n minus k other elements to be part of its complement. And so the way to form the subsets of size k is actually equivalent to counting their complements, which will be the subsets of n minus k. But of course, also, if you look at the definition here, not the definition, but the formula we have, n factorial over k factorial and n minus k factorial, be aware that n minus n minus k is the same thing as k. And this is multiplication, so the order does not matter. So if you look at how you choose k objects versus how you don't choose n minus k objects, that counts the exact same thing. So eight choose two is the same thing as eight choose six and eight choose three is the same thing as eight choose five and eight choose four is the same thing as eight choose four, which I confess is not super helpful, but it's important to recognize these things. Now binomial coefficients are not just useful for counting subsets. We can use these to answer lots of combinatorial questions because many questions in combinatorics are equivalent questions to counting subsets. And I should mention, of course, before I go too much further, that this notation n choose k, this is fairly universal in advanced mathematics, but some people in not advanced mathematics in earlier settings might use notation like n ck or c nk. This is sort of analogous to the following factorials we saw earlier in fall k. It could be denoted as n pk or p nk. Again, these notations are far less common in advanced mathematics, which is why we won't use them. We use the ones that we've introduced here. But be aware of this example you see on the screen here. How many different committees of three people can be formed from a pool of seven people? So let's say like we're trying to hire three new applicants from a pool of seven finalists. How many ways can you do that? If this is a set of seven people, what we're choosing is a subset of three people. And so the answer then would be seven choose three, which is seven factorial over three factorial and four factorial. So you get seven times six times five over three times two times one. You're left with seven times five, which is 35. You have 35 possible committees you could create from three people chosen from seven. All right, here's another example. And how many ways can a committee consisting of two faculty members and three students be formed if six faculty members and 10 students are eligible for the committee? Okay, so for example, maybe you have some students who work on the student senate or the student government and you have some faculty who work on the faculty senate or something like that. And you're gonna form some subcommittee. How could you do that? It's like, okay, we'll draft two faculty members from the senate. We'll draft three students from the student senate. How are you gonna do that? Now these decisions are actually independent of each other. So how many ways can you draft two faculty from a set of six? That'll be six choose two. But how can you draft the students? Well, the students you have to take three students here from the 10 students available, that's gonna give you 10 choose three in that situation. So then we work this thing out. Six choose two, six times five over two times it by 10 times nine times eight over three times two. We simplify these things. Two goes into six three times. So we get three times five for the first one. For the second one, two goes into 10 five times. Now I like the 10 there. Two goes into eight four times. Three goes into nine three times, like so. Three times five is 15. So that's how many ways you can select just the faculty members. And then three times four is 12 times 10 is 120. So the 120 is the number of ways that you can choose just the students. So if you're choosing faculty and students, the multiplicative principle says, since these are independent decisions, that you would have 1,800 possible committees you could get, 1,800 right there. And so these combinations we use to count the number of subsets of a set for a specific size. But choosing subsets, counting subsets is equivalent to so many other problems by forming committees. Remember here, combinations, the big thing about combinations is that they are unordered, just like a set, which is why these things count subsets after all.