 Welcome back to our lecture series Math 3120, Transition to Advanced Mathematics for students at Southern Utah University. As usual, I'll be a professor today, Dr. Andrew Missildine. And in the logic video for lecture 17, we introduced the technique, which we call combinatorial proof, which a combinatorial proof remember, you count the same set in two different ways. And if you do that, that means that the two formulas you come up with actually have to equal each other. I had mentioned that video that, while it's easy to understand a combinatorial proof, it can sometimes be difficult to actually come up with it. So I wanna give some more practice of combinatorial proof in lecture 18. Now I'm gonna break form a little bit. The typical standard we have for the lectures in this lecture series is the first video is about some type of mathematical topic. The second video is about some type of logical topic. And then our third video is about some type of communication topic. And admittedly the lines between these three topics are often blurred. Like sometimes we talk about language that has direct consequences to the logic or to the mathematics we've discussed previously in the lecture. Sometimes the language tips have nothing to do with what was in the other topics. Sometimes the proof techniques between the mathematical techniques are related. Sometimes they're not. And so in the previous lecture, number 17, when we introduced combinatorial proof in that lecture, the idea is these are proof techniques used in combatorics, which is what the math unit has been about for the last several lectures. So what we're gonna do in lecture 18 is we're gonna break form a little bit. We're gonna switch the directions. We're actually gonna do our logic video first, then our mathematical video. Because again, the topics are actually blurred together. In this video, we're gonna do some more examples of combinatorial proof. But I'm actually gonna focus all of my examples on proving identities about the binomial coefficients. Some of which we already know, some of which will be new to us, but all of them we're gonna prove using combinatorial proof. So remember, we're gonna count the same thing in two different ways. That's the strategy here. So the first one, this is an identity we've already seen before, but I wanna provide a combinatorial proof to the identity here. So for any natural numbers, n and k, the binomial coefficient n choose k is the same thing as n choose n minus k. Now, yes, if you use the formula n factorial over k factorial times n minus k factorial, then it's very easy to show this identity. So we've seen it before, but the reason I wanna go through this example is again to provide another example of combinatorial proof. This formula proof that we could use to prove these two things shows that they are the same, but it doesn't give us any reasons on why they're the same other than they just so happen to have the same formula. Like are they the equal to each other by some divine coincidence or is it in fact destiny that they have to be the same thing, right? The combinatorial proof, the advantage of them is that in addition to proving that two formulas are equal to each other, it gives us reasons, combinatorial reasons why they're the same because they're counting the same thing. I mean, this is in contrast to like mathematical induction which we'll start talking about this in the next unit starting with lecture 19, which induction will prove that things are true, but it doesn't really give us any intuition on why they're true. Often there's a combinatorial reason for it and that's what I wanna explore right now. So to prove that these two combinatorial symbols give us the same thing, consider the set X which is the numbers one, two, three, four, five up till n. So positive integers up to n. And so what we're gonna do is we're gonna count the number of subsets of X of size K. So that's the set we're gonna count the collection of subsets of X of cardinality K. Now by definition of the binomial coefficients, n choose K is the number of subsets. Now when I'm saying by definition, I'm not saying like the formula, the in factorial over et cetera, I'm saying we define the binomial coefficients to count the number of subsets or a problem similar to that, equivalent to that. But nonetheless, we can take that as the definition of our binomial coefficients. These things count subsets, n choose K is the number of subsets of drawn from a set of size K and each subset will have cardinality K. That's what those things are. We argued using like permutations and the division principle that you can write this with some factorials, et cetera. But the definition of this binomial coefficient is it counts combinations which is equivalent to counting subsets. So that's the first direction. The number of subsets is counted using the binomial coefficients. Now conversely, the number of ways of choosing a subset A inside of X is actually the same as the number of ways of choosing its complement. By choosing A, I'm actually also choosing who's not in A. And so conversely, if I choose who's not in A, then I'm actually choosing who's in A. But if A is an arbitrary subset of cardinality K, its complement will be an arbitrary subset of cardinality in minus K. So instead of counting the subsets of size K, I can count the subsets of size in minus K. But what are those? The number of subsets of cardinality in minus K will be in choose in minus K for the same reasons as above. And as such, that then shows that these two combinatorial symbols are equal to each other. So notice this proof is telling us that these two symbols are equal to each other. Even if I don't have a formula to compute these things, they're gonna be equal to each other. Which is kind of a really interesting and beautiful proof there. Like the proof is independent of the formula. I don't have to know how to compute these things, but I do know that they're the same thing. And again, this is the richness and beauty of combinatorial proofs. We know these things are true because of what they're counting, not necessarily because we know how to actually compute these things. We don't need to compute them to know that these things are in fact one of the same thing. And this proof I've actually given to us previously in this lecture series. I intuitively mentioned this proof, but I've now actually put it into a more rigorous form. All right, let's now consider perhaps the most important binomial identity that exists. And we'll actually use this to develop Pascal's Triangle in the very next video, which is why we're switching the order of our logic video with our math video. We're doing it second so we can use this identity to help us understand Pascal's Triangle and the binomial theorem. But we're gonna prove this identity using the combinatorial proof method. So take two natural numbers, N and K here. And so I claim that the binomial coefficient N plus one choose K is the same as the sum of binomial coefficients N choose K minus one plus N choose K. And again, we don't actually have to have a formula of these things to prove this thing is true. We're gonna prove it by counting the same collection of things in two different ways. And so this proof is a formula-independent proof for which if you try to do this with formulas, you can do it, but it's gonna get messy and awkward. And if you are successful in proving that the left-hand side equals the right-hand side, your proof basically will feel like a trig identity. It gives you, you can prove it's true, but it might hide. It might obscure why it's true. The combinatorial proof puts it out in front. This is true because we're counting the same object. Now we're gonna make a slight modification to the counting we were doing from the previous theorem. This time, the set X is giving me the number of elements one, two, three, four, up to N plus one. So we're gonna consider integers up to N plus one positive integers. And that's to coincide with this thing is counting subsets of a set of cardinality N plus one. Oftentimes with these combinatorial proofs, one side of the equation will be generally simpler than the other and it's almost screaming to you what it wants to count. So N plus one, choose K, counts the number of subsets of a set of N plus one, which notice X has cardinality N plus one. And so if we wanna count the subsets of cardinality K, that is gonna be N plus one, choose K. Sorry about the typo right there. And that gives us, of course, the left-hand side of our formula. Again, sorry about that. So that gives us the low hanging fruit. Now what we have to do is look at the right-hand side. And so let me give you some idea of what we're doing here. Binomial coefficients count subsets. That's what you wanna think about when you think of binomial coefficients. You can think about counting combinations, that's fine as well, but binomial coefficients count subsets. That's what we started with the left-hand side, counting subsets of this set with N plus one objects. What does plus mean in terms of combinatorics? So our formulas are gonna involve our arithmetic operations, addition, subtraction, multiplication, division. We've learned all about these four counting principles. And so when we see arithmetic operations like addition, subtraction, multiplication, division, that tells to us some type of combinatorial structure. The addition principle tells us that we have some type of partition in play. When we have two mutually exclusive cases, if you consider those cases and these cases, and since they're mutually exclusive, their sum will give you the union of those two cases. And if there's only two cases, if this is exhaustive and mutually exclusive, that sum suggests to me that I wanna take the set from before and I wanna break it up into two cases, this or that, either or situation, two cases that are mutually exclusive. So I was counting subsets before, so I wanna count subsets of a set of size N with K minus one elements versus a set of size N with K many elements. So that intuition is why I then will proceed in the following manner here. So for the right-hand side of the equation, whoops, for the right-hand side of the equation, we're gonna organize the subsets of X based upon does the subset contain the element N plus one or does it not contain the element N plus one? This is the either or situation we're gonna do and these are mutually exclusive. Either the subset contains N plus one. If it does, it goes into this category. If it doesn't contain N plus one, it's a separate case, we're gonna consider that separately, okay? So we know that our sets A, we're choosing sets of cardinality of K, okay? That was the original problem. Now, if A contains N plus one, that accounts for one of the K many elements in A, then what's left over is K minus one unaccounted for elements so far. So we'll let the set B be the set A where we've taken away the distinguished element N plus one. And likewise, I'm gonna let the set Y be the set one, two, three, four up to N. Notice that Y here is just X take away N plus one as well, okay? So B is a subset of Y because B is the subset of A that doesn't contain N plus one anymore. So B is a subset of Y and because A contained K elements, B is gonna have to contain K minus one elements. So as we allow the set B to vary, there are in choose K minus one possible ways to choose the subset B from the set Y because again, B contains K minus one many elements. Now, if I would choose a random B in this format, I can construct a set A by just taking B union that missing element N plus one. So I can reconstruct A by this subset B as well. So as we allow A to range over all the possible subsets of X that do contain N plus one, this is the same counting problem as choosing an arbitrary subset of Y which won't have N plus one in it. And again, since A has K elements, B will have K minus one elements. So there are in choose K minus one ways of choosing the set B, which then means there are in choose K minus one ways of choosing the set A where A will be a subset of X with K elements and it will contain N plus one. So basically there's in choose K minus one ways to choose the other K minus one elements of A other than N plus one. So that is one of, that's one possibility. If A contains N plus one, there's gonna be in choose K minus one ways to do it. Now, conversely, if A doesn't contain N plus one, then it's still a subset of X that contains K elements, but if it doesn't contain N plus one, then it turns out that A is a subset of Y and like I said, it still contains K elements. How many subsets of Y can we choose? Well, there are in choose K many subsets of Y that have cardinality K and since A doesn't contain N plus one, we can identify it with one of those subsets. Now remember, A either contains N plus one for which we get this possibility or it doesn't contain N plus one for which we contain this many possibilities. These two cases are mutually exclusive. You can't contain N plus one and not contain it and it's also exhaustive. It's an either or situation. So by the additive principle, which I predicted earlier, by the additive principle, we can add together these two cases because every case is one or the other and not both and therefore the number of ways to choose a subset of X of cardinality K is the sum of these two binomial coefficients and then by combinatorial proof, those two combinatorial symbols have to be the same thing. N plus one choose K is equal to N choose K minus one plus N choose K and in the next video, we're gonna explore how we can use this identity to actually compute the binomial coefficients recursively. That'll be a very important term once we get into our next unit about integers and mathematical induction but I'll address that of course in the next lecture. In this video, I do wanna still continue and do some more examples of combinatorial proof, proving some identities for the binomial coefficients. So this is a nice one right here. Let N be a natural number. And again, this does include the possibility that N is equal to zero. That is true. These statements are all true with N equals zero or K equals zero as well. Then if you take the sum of all of the binomial coefficients where the top number is always N, you get N choose zero plus N choose one plus N choose two plus N choose three all the way up to N choose N. This is the same thing as two to the N. And so the left-hand side, we could simplify this as you take the sum where K equals zero to N and then you take the sum of the binomial coefficients and choose K. So that's what we're doing right now. That we claim this is equal to two to the N. So when I look at this formula, I wanna prove it by combinatorial proof and the formula actually suggests to me the proof we're gonna use. So the first thing I notice is the right-hand side looks like two to the N. Two to the N, I feel like I've seen that number before and sure enough, two to the N was the cardinality of a power set. And so that's actually what I'm gonna try to count. I'm gonna try to count the power set of a set in two different ways. So if the set X contains N elements, let P of X denote the power set of X. Remember, this is the family, the set of all subsets of X, including the empty set and the set X itself. And so we proved previously in our lecture series that in corollary 143, the cardinality of the power set if X has itself cardinality N will be two to the N. So in one respect, the cardinality of the power set is this exponential two to the N. So what I wanna then do is show that this number right here can be computed as the sum of those binomial coefficients. But then like we said with the previous example, I'm adding together terms. I want to use the additive principle here, which means I need to consider a partition. I need to break up the power set X into a collection of non-overlapping possibilities so that every possibility is considered. And so that's what I'm gonna do with these additions, but then each symbol that I'm adding is a binomial coefficient. It counts subsets of a fixed size. And so that suggests to me how I'm gonna approve this thing to be true. I'm gonna take all of the subsets and partition them by their cardinality. We'll put all the zero sets together, all the one sets together, all the two sets together, all the three sets together, all the four sets together. If we organize the subsets of X by their cardinality, then I can use the additive principle to help me out here. So note that the number of subsets of cardinality zero inside of X, that number is N choose zero. There's only one of them, the empty set, but nonetheless, that statement is still true. How many singletons do we have? Sets, subsets of cardinality one. Well, N choose one counts them. How many subsets of cardinality two? Well, that's N choose two. And then if you keep on doing this, N choose three will give us all the subsets of cardinality three, N choose four will give us the cardinality, sorry, N choose four will count the number of subsets of cardinality four up until we get to N choose N, which will give us all the subsets of cardinality N, which is of course X itself. All right, now every subset has a cardinality and therefore every subset will be counted once and only once in one of these cells. We do in fact have a partition in hand here. And therefore by the additive principle of counting, if we add together the possibilities of all of the cells in the partition, there's no overlap, so we don't need to use inclusion exclusion right here. We get the sum of all the binomial coefficients will equal the cardinality of the power set. And as we've previously shown that that power set is equal to two to the N, this shows us that the sum of all the binomial coefficients will always give us a power of two where the power of two is this number here on the top. It's super, super nice, right? Again, I don't need to know how to compute a binomial coefficient to prove this formula. I can prove that the sum of them is gonna add up to be two to the N because of by virtue of what they are, they count subsets and as there's two to the N subsets, the sum of the binomial coefficients will always be a power of two. Do some examples yourself to verify it. This is a valid proof. And we did this combinatorially. All right, I'm saving the best for last and best in this case is mostly the complexity of this thing. This is gonna be the most complicated formula of this video but I promise this proof by combinatorial methods is exactly what we're gonna do here. Let's prove the identity that for any natural numbers N and K, which again, the N and K could be zero, right? Then I should also mention that if N happens to be smaller than K and all these examples we've considered so far, the associated binomial coefficients would actually be zero. Like if you have like three elements and you wanna choose five, how many subsets can you form? There's zero though. So if K is larger than N or if K is negative or something like that, which of course that's not the case if it's a natural number but some of these binomial coefficients are zero. So those exceptional cases, these formulas are also valid in that situation. I haven't been drawing much attention to that but this is valid for any natural numbers here. So what I wanna prove this time is that the sum as K ranges from zero to N of the binomial coefficients N choose K squared is equal to the binomial coefficient to N choose N in that situation. Now the left hand side is far more complicated in this example than the right hand side. And so I'm actually gonna use the low hanging fruit to determine what object am I gonna count in two different ways. The right hand side is a single binomial coefficient. What do binomial coefficients do? They count the number of subsets. So what I do want is I want a set whose cardinality is two N. And so we're gonna take X to be equal one, two, three, four up to two N. And this will give us a set whose cardinality is two N. So since the cardinality of X is two N, the number of subsets of cardinality N is going to be, oh, I'm sorry, that wasn't a typo. I didn't read my sentence. This is the problem. If we have X equals the set one through two N, then the counting problem we want to consider is to count the subsets of X whose cardinality is N. So we wanna count those subsets whose cardinality is exactly half of X. Well, since X's cardinality is two to the N, the right hand side, this binomial coefficient two N choose N is exactly what would count that. By definition, these binomial coefficients count the number of subsets of X whose cardinality is N. So that gives us the right hand side. So now we need to investigate the left hand side and see how does this count the subsets of X whose cardinality is N. Now, the structure, the arithmetic structure of the formula is actually gonna suggest how we proceed forward. There is a sum. So like the previous examples, I probably need to come up with a partition. I need to consider separate cases that don't overlap but are exhaustive. A partition is what I have in mind so I can use the additive principle. But notice that you're squaring something here. You have N choose K squared. Now that's gonna actually use the multiplicative principle because we multiply things together. If you have independent choices, then the products, the possibilities turn out to be list inside of a Cartesian product. We multiply them together. So I'm adding together products so I probably want to come up with a partition for which each element in the partition is itself two independent choices. So that's how I'm gonna try to break this thing apart. And so in order to do that, what I'm going to do is I'm going to select an arbitrary subset A of cardinality N. So I'm just gonna choose one. So this is just, I'm just gonna pick one subset of cardinality N and boom, it's done. It's gonna be fixed for the rest of the problem. I don't care which one it is but just choose one of them. It's a subset of X, its cardinality is N. So this is one of the sets we're trying to count. It's a distinguished one. Much like earlier we had this distinguished element N plus one, we got this distinguished subset of A. All right, now for simplicity and formula, I'm gonna let B be the complement of A. So this is everything that's in X that's not in A. In particular, you'll notice that A union X is equal to X and these sets don't overlap. This is a partition. Both of these cells are, both A and B have cardinality in itself. Now this is not the partition we're gonna use because the previous formula, this is a big sum. There could be lots of terms in the sum. So this partition will help me get the partition I'm looking for. So what we're gonna do is now with this set A and B in hand, we're gonna consider what happens as K ranges from zero to N. So what I want us to do is then look at subsets. So if we take an arbitrary subset of X who itself has cardinality N, that set is gonna have some objects that belong to A and it's gonna have some objects that belong to B. So like if we just take an arbitrary subset, let's call it, let's call it C, okay? C is a subset of X. Well, C, if I look at its intersection with A, there's gonna be some elements that belong to C and A. Potentially, I mean it could be zero, right? But then also consider C intersect B, the complement of A. It's like everything that belongs to C either belongs to C intersect A or belongs to C intersect B. So if I take the union of this, this is gonna give back C, much in the same way that A union B gives me back X. And so given the partition A union B on X, I can form a partition of C with respect to A and B as well. So every element of C belongs to C intersect A or belongs to C intersect B. In other words, every element of C belongs to A or belongs to B because everything in X belongs to A or B. And so let's consider the possibility where K objects of C belong to A and hence in minus K objects belong to B because after all, this set C has a cardinality of C. This is one of the typical elements we're considering right here. And this set C is arbitrary, right? So A is now fixed. It was arbitrary, but we fixed it. That is where it's not gonna vary anymore. C is a potential one. If we take just a typical set which shares K elements with A and hence contains, it shares in minus K elements with B then we can form C by choosing randomly K elements from A. There's in choose K ways to do that. And we could randomly choose in choose in minus K elements from B. And so we could randomly generate these sets C. It's like, okay, I'm just gonna take my elements from A and put them into C. There's in choose K ways to do that. And then I'm gonna take in minus K elements to put into C. So there's gonna be in choose in minus K elements to do it that way. And so by the multiplication principle there are gonna be in choose K times in choose in minus K ways of forming such a subset of X. Where again, C has a cardinality of N, but the intersection between A and C has a cardinality of K. So for that, for this K right now there's gonna be in choose K ways to choose the elements that live inside of this set. And then there'll be in choose in minus K ways to choose the elements that live inside of this set. So every subset of X with cardinality N whose intersection with A is K would be formed in such a way. So there's this many possibilities from the multiplication principle. All right, so then again, let's allow K to range as K can range from zero because the intersection between C and A could be zero all the way up to N because C could be A itself. That's a possibility. As you allow for those, I want you to be aware that these are now mutually exclusive cases. If this one set has two elements in common with A and have a different set which has three elements in common that can't be the same set. So I'm not counting it twice. So as we allow K to range from zero to N this then forms a partition of all of the subsets of X whose cardinality is N based upon how large their overlap is with A. And so by the additive principle we then can add all of these possibilities together and we get the following. We get the sum where K ranges from zero to N because those are the possibilities where we get N choose K times N choose N minus K. So if you break out that sum there's this case where you have zero elements in A and N elements in B. There's the case where you have one element in A and N minus one elements in B. There's the case where you have two elements in A and hence N minus two elements in B all the way down to you have all N elements in A and no elements in B. So those are all the possibilities there. And so by common editorial proof this sum is equal to the in the two N choose N that we had considered previously. So we get this identity by common editorial proof but wait a second. N choose N minus K is the same thing as N choose K. We prove that as the very first thing in this lecture video. So that was theorem three, 10, three so that this can actually substitute with that. And then you get N choose K twice and so you end up with the formula we have right there. And so that then brings us to the end of this proof and that's also gonna bring us to the end of this video. So we're gonna solve four fantastic common toadal proofs to prove identities about binomial coefficients. So what I want you to get from this is the basic idea of common toadal proof is to count the same set in two different ways. And the formulas themselves can actually be suggestive of how the formulas, how you prove them, right? Binomial coefficients count subsets. So we start coming up with sets and we wanna count their subsets. Multiplication means we could use the multiplicative principle. So we're gonna have list. We have decisions, list of decisions that have to be done in order. Some represent partitions. And so that we didn't do any with the division principle or the subtraction principle, but when you see these divisions, when you see these symbols inside of the formulas, they tell you which counting principles to use and that often helps us learn how to prove it combinatorially.