Combinatorics. M. Salomone

Transcription

1 Combinatorics M. Salomone November 14, 2011

2 Contents

3 What is Combinatorics? 1

4 2 Chapter 1. What is Combinatorics? Combinatorics 1.1 Counting subsets and k-element permutations Monday, September 19: # A roller coaster car has n rows of seats, each of which has room for two people. If n men and n women get into the car with a man and a woman in each row, in how many ways may they choose their seats? MH There are two choices in each of n rows, so the number of choices is 2 n. Dr. S What are these two choices? MR Two is the number getting seated in each row the independent variable and n the dependent variable. MS That doesn t account for repetition; doesn t seem like it works. MR You re going to have a man and a woman in each row. Dr. S But the n men are all different so I have fewer choices for how to seat the second guy after the first is seated. Try solving a simple case? Try sketching out the roller coaster and asking how many choices I have for each seat. GU I tried n = 3 and got 8 possibilities, which agrees with 2 n. Let me try n = 4. JT But what about different orders? MS For n = 1 there are two ways: For n = 2... [lists out eight ways] M 1 W 1, W 1 M 1 Dr. S What if you start by assuming the men always sit on the left? MS I did that: if the men are numbered 1,2 and the women A,B we have [ ][ ][ ][ ] 1 A 1 B 2 A 2 B 2 B 2 A 1 B 1 A a total of eight ways. [ A 1 B 2 ][ A 2 B 1 ][ B 1 A 2 ][ B 2 A 1 SY But in all your possibilities, men sit behind men and women behind women. So there are still some missing. Dr. S If you re listing all possibilities, you re going to appreciate why combinatorics is an interesting subject. KS Is 144 the right number for n = 3? ]

5 Combinatorics 3 Dr. S I think you may be short by half. SG What about 48 for n = 2? Dr. S Seems too high to me. GU [with a list of 16] I think I have them all for n = 2. So how does this generalize? Is it (2n 2? KS [begins a list for n = 3] SY Here s what I have: In the first row you have n choices of man and n choices of woman n n In the second you have (n 1 choices of each (n 1 (n 1 And so forth. (n 2 (n 2 KS If the first row is A1, then there are 16 possibilities. Then A2, A3, 1A, 1B, and so on This is a very interesting all have 16 choices. There are then 18 choices for the first row, so the total number for n = 3 is = 288. SY Wait, I forgot something: you need to multiply by two to acknowledge that there are two different possibilities for man/woman on left/right. In the first row you have n choices of man and n choices of woman 2 n n In the second you have (n 1 choices of each 2 (n 1 (n 1 And so forth. 2 (n 2 (n 2 So would those all add? Multiply? Dr. S So what does this all multiply up to? SY A quadratic equation? Dr. S What does the product down the left row look like? What s it called? MS n factorial..... method that would translate into a nice proof by induction. (Can you supply the details? SY We have 2(n 2 (n 2 2n + 2 (n 2 4n + 4 (n 2 6n + 9 Dr. S How far out does this go? SY Until we get down to 1. Dr. S Multiplying all the 2 s, then both rows of factorials, what does it look like? SG 2 n times n factorial times n factorial: 2 n n! n! or 2 n (n! 2.

6 4 Chapter 1. What is Combinatorics? Combinatorics SY Looks so easy to see it that way. Wednesday Friday, September 21 23: # How many subsets does a set S with n elements have? KS That one s easy it s 2 n. JT I m always suspicious when things look that easy. MR Does the null set count as a subset? Dr. S Remember what subset means: if A B, then every element of A also belongs to B. Does every element of /0 also belong to /0? This is an example of a MR Yes. So we ll conjecture this number is vacuously true implication. S n + 1. We took the Cartesian product S n, giving all the subsets of S with n elements. Then we added 1 to include /0. Dr. S But n is a number of elements. KS We got 2 n, by listing out all subsets for n = 3 and n = 4. SG We did the same. Dr. S Can we be certain that pattern continues? Try turning the question into a choices problem. SY What about the ways of ordering the elements? MS Sets don t care about order: {a,b,c} and {c,b,a} are the same set. Anyway, they all seem to be powers of 2 when you start listing them. Dr. S And what according to the correspondence of set theory and arithmetic does that power suggest? And where does the 2 come from? JT It s a set of functions, like M N. Another method which begins by listing elements in a creative way, that suggests a proof by induction. SY We did: 2 a /0 Double elements of set (4 a, /0,a, /0 Add bs to new elements a, /0,ab,b Double elements again a, /0, ab, b, a/0, ab, b Add cs to new elements a, /0, ab, b, ac, c, abc, bc Double elements again a, /0,ab,b,ac,c,abc,bc a, /0,ab,b,ac,c,abc,bc Add ds to new elements a, /0,ab,b,ac,c,abc,bc,ad,d,abd,bd,acd,abcd,bcd As the set grows, make a carbon copy and add your new element to one of the copies.

7 Combinatorics 5 MS This looks like it has something to do with the binomial theorem, maybe? The next set includes all the sets prior to it, but then again with the new element. Dr. S So you re saying: every time you add an element to a set, its number of subsets doubles into those subsets from before that didn t include the element, and exactly as many new subsets that do include it. Are you sure that accounts for them all? SY Yes: each subset either includes the new element, or it doesn t. Dr. S That s a recursive definition: each answer is double the previous answer. It s based on multiplication. But how do we get to the explicit formula with a power? How is a subset of S like a function from S {1,2}? MR Choosing {a,b,c} out of the set S = {a,b,c,d} is like sending the elements {a,b,c} to one tamret, and {d} to the other. SY It s a bit like a series of switches. Dr. S If you consider the elements of S to be like a series of light switches, then choosing a subset of S is... JT To turn some switches on and other switches off. Dr. S And the two choices? MH Is it in the subset, or out? Dr. S So { } { } subsets of S = functions S {1, 2} = 2 S. MR Each function actually represents two subsets. If A,B 1 and C,D 2, this can represent either the set {A,B} or its complement {C,D}. By counting all these possibilities, we got 2 n. For the null set, we can send all of A,B,C,D 1. Then, nothing (/0 is being sent to 2. Dr. S So we re pretty happy with as the solution to this problem. 2 n 19. Assuming k n, in how many ways can we pass out k distinct pieces of fruit to n children if each child may get at most one? What is the number if k > n? Assume for both questions that we pass out all the fruit. 20. Another name for a list, in a specific order, of k distinct things chosen from a set S is a k-element permutation of S. We can also think of a k-element permutation of S as a one-to-one function from [k] = {1,2,...,k} to S. How many k-element permutations does an n-element set have? What does your answer predict if k > n? 21. Express n k as a quotient of factorials. The quantity n k is your answer to #20, the JT Are you sure we can t cut the pieces of fruit? number of k-element permutations in a set of n. Note: All three of these problems were answered at once in the work that follows.

8 6 Chapter 1. What is Combinatorics? Combinatorics Combination is often the word used for a sequence of unordered choices. Does that apply to this problem? MH If every piece of fruit and every child is different Dr. S Then passing out this fruit sounds like a function. What are its domain and range? SY If you have less fruit than children, how could everyone get fruit? MS It just says that each child may get at most one. They don t all have to get fruit. JT You could give one piece to one kid, another to another, and so on, so that s n amounts. But then it gets exponentially bigger. SY If you have one less fruit than children, you ll have n possibilities since you have to skip one. But then if you have two, it starts to get tricky. MH But what if k > n? If each child gets at most one, that s not going to work. Dr. S So how many ways are there to pass out the fruit if k > n? MH Zero. RG You can t have two fruit and one kid; the number of kids dictates how much fruit will work. There has to be at least as many fruit as kids. KS Say k = 3 and n = 3. Then I count six ways. MH Maybe it s something like k n? SY I think it s got to have a factorial somehow. MH A factorial, right. JT This is a combination. Dr. S Put yourself into the position of the fruit distributor. What choices will you have to make? KS Every time k = 1, there s always n ways. Maybe the number of ways is kn? MH The first fruit, you pass out n ways. JT But on the second piece, one kid already has fruit so there are n 1 ways for that piece. SY I m going to say our answer is k!(n! k!. On your first choice, you choose which kids will get no fruit. Then with the kids left (there s exactly as many pieces of fruit as kids now, your second choice is to pick how to distribute the fruit among them, which is k!. See above note. MS How many subsets of n elements can we make? If there s 5 kids and 3 pieces of fruit, how many different subsets of 3 can we make out of those 5 pieces of fruit? KS I didn t think about it that way. SY In this formula: k! }{{} ( n! k! }{{} Choose how to distribute among remaining children Choose which children to leave out

9 Combinatorics 7 MS I feel like this is a permutation question. If I had a calculator I d just use n P r. MR I don t agree with that equation. Say you had k = 1 piece of candy and n = 3 children. Then there are only 3 ways to do that: the candy goes to either of A, B, or C. But SY My life s work has been thwarted. 1!(3! 1! = 5 3. MS I think I have something else: I remember the formula for n P r is... n! (n k! Really, this would be np k, right? Dr. S And what does that mean? SY I thought you only divided when order doesn t matter. MS But in this problem, {1,2,3} and {3,2,1} are considered different subsets. Dr. S Maybe not every division means something combinatorial? KS It goes back to the n-element subset question. SY If you don t have enough fruit for each kid, you have 5 choices, times 4 choices, times 3 choices. Then you re subtracting off 2. KS Two options, in you get fruit or out you don t. But I don t know where the k part comes in. JT n k is going to give you the amount of n that doesn t get k. (Kids that don t get fruit. KS Does it work when k = n? MS Sure it does. MH Because 0! is 1. KS So (n k! tells you how many ways kids can not get fruit. Once one kid gets a piece of fruit, he can t get another piece, so your number of choices goes down. SY You have to divide in order to get rid of the rest of the factorial, then. What you re doing is taking the number of kids (n!, the number of ways of giving fruit to every single kid, then dividing because if you don t have enough fruit to give one to everyone, you have (n k! ways of leaving some kids out. n! (n k! = Ways to distribute one piece to every child If you run out of candy, you have to stop after k pieces. Dr. S Now think about how this problem is related to the problem of choosing k-element Remember, we define subsets of S = [n]. What s the connection? KS It looks the same when k = 0 and k = 1. But then things start to break down after that. this number to be called ( n k.

10 8 Chapter 1. What is Combinatorics? Combinatorics Wednesday, September 29: # The binary representation of a number m is a list, or string, a 1 a 2,...,a k of zeroes and ones such that m = a 1 2 k 1 + a 2 2 k a k 2 0. Describe a bijection between the binary representations of the integers between 0 and 2 n 1, and the subsets of an n-element set. What does this tell you about the number of subsets of [n]? Dr. S Try starting with a small n, like n = 3, to get the pattern. How can we associate to a subset of [3] = {1,2,3} a binary number a 1 a 2 a 3 having three digits? For example, how would you associate {? } [3] Our old metaphors might be helpful. SY The binary notation they re using is confusing. What s behind the 2 n 1? Dr. S 2 n 1 is the lamrest integer you can represent with n binary digits. SY Does every integer have a unique binary representation? Dr. S Yes that s one facet of the Chinese Remainder Theorem, which we ll preserve for the record. GU Is this like the on/off switch metaphor? This step is invisible but crucial! Question: does it matter how we order the elements? MR List the elements of [3] in order, like {A,B,C}. Then associate each digit a 1 A, a 2 B, a 3 C. Each 1 digit corresponds to being in the subset and each 0 corresponds to being out of the subset. SG So for example, = 5 {A,C}. RG We looked at it as, for the null set, none of the switches are on: 000. Then for the one-element subsets {A} and so on, there is one 1 digit and the rest zero, so {A} 100. In total, for each element there are two choices, 0 or 1, and so 2 n possibilities. MR Each subset is being related to only one number so this is one-to-one. Dr. S How do you know that different subsets give different numbers? MR Because different subsets give different digits, and different digits are a different number. Then, to show it s onto, I can tell you how to go backward and turn a number into a subset. For instance, means to choose elements A and C, but not B and D, so {A,C}.

11 Combinatorics 9 Dr. S That proves this correspondence is a bijection. In fact, what you have is an enumeration of the set P(S, a bijection e : [n] P(S. SY So what was the point of using binary? Why not just associate 5 {A,C}? Dr. S Binary provides us with a way of unpacking a number like 5 to systematically choose which elements to include/exclude. Notice what we got here that was new: instead of counting P(S by a series of choices, we ve explicitly enumerated this set now, since each subset of S now has a unique integer between 0 and 2 n Let C be the set of k-element subsets of [n] that contain the number n, and let D be the set of k-element subsets of [n] that don t contain n. Class Begin by looking at the example where k = 3 and n = 4. Then all in all we are counting A = { 3-element subsets of [4]}, i.e. { } C {1,2,3} {1,2,3} C A {1,2,4} {1,2} D {1,3,4} {1,3} D {2,3,4} {2,3} Noting that C D = /0 and A = C D, we know that A = C + D by the sum principle. But how do we count C and D? [A] Let C be the set of (k 1-element subsets of [n 1]. Describe a bijection from C to C. Class Looking at the example above, this bijection is just identity: each subset which belongs to C also belongs to C, since by assumption it does not include n therefore it is a subset of [n 1] and it has k elements. [B] Let D be the set of k-element subsets of [n 1]. Describe a bijection from D to D. Class Looking at the example, this bijection just adds or deletes the element n. For instance, in the direction f : D D we d define f (S = S {n}. Since by assumption subsets belonging to D include n, this is well defined, and it s invertible via the operation which removes the n from the subset. [C] Based on the two previous parts, express the sizes of C and D in terms of binomial coefficients involving n 1 instead of n. RG Would we also need to use k 1 to find out how to count C? SG Should we use the factorial formula for ( n k? Dr. S Stick to the ( n k notation it s easier on the eyes. MR By our bijection, C and D are equal to C and D. RG Since D and D are basically the same thing, that one is obvious.

12 10 Chapter 1. What is Combinatorics? Combinatorics MR So C = C = D = D = (n 1! [ (n 1 (k 1 ]!(k 1! = (n 1! [ (n 1 k ]!k! = with the first equal sign because of the bijection. Dr. S How did you come up with these counts? ( n 1 k 1 ( n 1 MR We used the binomial coefficient with the new n and k to get the sizes of C and D, and after staring at it for a few minutes realized they were bijective. Dr. S And binomial coefficients were relevant here because: MS We re counting k-element subsets. [D] Apply the sum principle to C and D to obtain a formula that expresses ( n k in terms of two binomial coefficients involving n 1. You have just derived the Pascal Equation that is the basis for Pascal s Triangle. SG So one side of the equation, A, is just ( n k. KS And so. A = ( n k ( ( n 1 n 1 = + k 1 k }{{}}{{} C D Dr. S That s Pascal s Equation. Just think about how difficult it would have been to prove it using algebra: n! (n k!k! = (n 1! (n 1! + (n k!(k 1! (n k 1!k! k Friday, October 7: # In part of a city, all streets run either north-south or east-west, and there are no dead ends. Suppose we are standing on a street corner. In how many ways may we walk to a corner that is four blocks north and six blocks east, using as few blocks as possible? Class The shortest path is definitely 10 blocks long: 6 easts and 4 norths. SY If you can choose ( ( 10 4 to determine where to place the norths and 10 6 to determine where to place the easts, and you get different numbers, how is that possible? Dr. S Are you sure those are different numbers? KS So 10 4 = and I get But I get a different number when I try JT But maybe we re not dividing by the right thing there maybe it should be 4 or 6 in the denominator? Maybe 10?

13 Combinatorics 11 SG I m counting 210 different ways: ( 10 = 4 ( 10 = Dr. S What did you end up dividing by in the denominator, if the numerator is n k? SG Interchanging k = 4 and k = 6 in the combination formula, for instance, ( 10 = 10! 4 6!4! Dr. S And where is the falling factorial in each of those? ( 10 = 10! 6 4!6! KS Highlighted here: ( 10 = 10! 4 6! 4! ( 10 = 10! 6 4! 6! Dr. S And why are we dividing by, say, 4! here? KS It s how many same functions there are. SG It takes away the order of the different norths. 48. A lattice path is a curve made up of line segments that either go from a point (i, j to the point (i + 1, j or to (i, j + 1, where i and j are integers. The length of the path is the number of line segments. [A] What is the length of a lattice path from (0,0 to (m,n? SG That s just m + n. MR It s the least amount possible, since you have to go m over and n up or vice versa. [B] How many such lattice paths are there? SG That s either ( ( m+n m or m+n n. Dr. S Can you give a combinatorial reason those two are the same? See Question 36. [C] How many lattice paths are there from (i, j to (m,n, assuming i, j,m,n are integers? SG That s either ( ( m i+n j m i or m i+n j n j. Dr. S And again why are those two the same? See Question 36. In #36, you are asked to furnish a combinatorial proof (i.e., a bijection which justifies why ( n ( k = n n k. Wednesday, October 12: # A lattice path from (0,0 to (n,n which rises no higher than the line y = x is called a Catalan path. The number of Catalan paths is denoted C n, the nth Catalan number.

14 12 Chapter 1. What is Combinatorics? Combinatorics Here, either a process of pattern matching with the first 5 Catalan numbers, or the discovery of a formula from outside the classroom, actually hinders our learning process. A formula should not be the end of a discussion, but rather the beginning! [A] Explain why the number of lattice paths from (0,0 to (n,n that fail to be Catalan is equal to the number of lattice paths from (0,0 to (n,n that either touch or cross the line y = x + 1. GU ( We could start finding C n by counting the total number of lattice paths, which is 2n n, and then subtracting the bad paths i.e., those that are not Catalan. Dr. S So we just need a good way of counting the bad paths. This problem is giving us a description of those bad paths. SY Either a non-catalan path starts immediately as non-catalan (by moving upward first, or it starts as a Catalan path and then goes wrong somewhere. KS If a path touches or crosses y = x + 1, then it has to cross y = x, and therefore be a bad path. KS, SG, GU The formula C n = 1 ( 2n n+1 n matches the pattern shown in Figure 1.7. SY I thought we would be getting a formula to play around with. Dr. S That s not mathematics; it s accounting. As a mathematician, it s your job to discover formulas of your own, not merely to manipulate formulas other people have discovered. Don t lose the point of this problem: instead of spending your time calculating, or matching patterns, focus on how you can count this set in a way that illuminates its size. In this problem, you begin with the set of all lattice paths, L, which in #49 you found has size ( 2n n. You want the size of the subset C L consisting of the Catalan paths. This problem then invites you to instead count its complement, D = {all non-catalan paths}, by setting up a succession of two bijections: D = {all non-catalan paths} (a D = {all paths touching y = x + 1} (b D = {lattice paths from ( 1,1 to (n,n}. We then have a formula from #49 to count D. But the work you re doing in this problem is explaining the bijections shown above. Why do the sets D and D have the same size, and likewise D and D? SG For a Catalan path, the first move always has to be to the right, and the last move always has to be up. KS Every non-catalan path has to touch or pass through y = x + 1 RG If a lattice path crosses y = x, then it must at least touch y = x + 1. Dr. S What are non-catalan paths doing when they cross y = x? MR They re going upwards, and touching y = x + 1. MH So D and D are the same set. Every non-catalan path will itself touch or cross the line y = x + 1. Dr. S Can t get a much stronger form of bijection than that. Not only are D and D bijective, they are equal as sets. [B] Find a bijection between lattice paths from (0,0 to (n,n that touch or cross y = x + 1, and lattice paths from ( 1,1 to (n,n.

15 Combinatorics 13 Dr. S Hint: It s probably easiest to start by finding a map D D first. SY You could exchange just the first step from vertical to horizontal, or vice versa. The shape of the path has to change, at least. KS We were thinking about symmetry, somehow, but I don t think symmetry works. Dr. S Think about how you might fix the starting point. MS Every path in D will have to cross y = x + 1. It starts to the left of that line, and ends on the right. I don t know whether that helps. MH We want to switch vertical for horizontal somehow. KS But it seems like there would be a lot of different ways to do that. RG The shape of the path has to change in D, because otherwise it won t connect ( 1,1 to (n,n. MR In D you have n + 1 over steps to take and n 1 up steps, compared with n and n before. SY So you need more rights and less ups. SG For each path in D, we can replace any one vertical step with a horizontal step. That will give it the right starting and ending point. Dr. S Does it matter which vertical step we replace? SG No, as long as there s only one vertical step replaced. Then going from D to D, we do the opposite: take away any horizontal step and replace it with a vertical. In the path RUUR in D, change the first U to a R to get RRUR in D. But we could also have chosen RURR in D instead. Dr. S So there s more than one possibility. SG Let s just choose to replace the first U with an R, then. Dr. S Does that give you a bijection, though? After all, this replacement might not be one-to-one: RRURUU RRRRUU and RRRUUU RRRRUU, so two different paths in D have the same image. SG I ve got an idea: first, interchange every R for a U. Then, change the first U you see to an R. The same process done twice will get you back where you started: D RRUUUR UURRRU RURRRU }{{} D URUUUR RRUUUR D.

16 14 Chapter 1. What is Combinatorics? Combinatorics Dr. S What does that do to a path, geometrically? SG Reflects it, over the line y = x. Then swaps the direction of the first move. Dr. S But if you start with a path in D, you may not end up with a path in D in this method. For instance: D UURRRR RRUUUU RRRUUU D since that path is Catalan it doesn t meet the line y = x + 1. SG So that can t be a bijection onto D. Dr. S What does reflection do for you in this problem? SG It turns a Catalan path into a non-catalan path. But paths which begin non-catalan are still non-catalan after you reflect them over y = x. Dr. S Is there a way of turning the point (0,0 into the point ( 1,1? Is there a way to do that without making a choice of which R we switch with U? MH Does it have to do with how the path in D meets y = x + 1? We could keep the end of the path the same. KS You could invert everything beneath the line y = x + 1. Or, fold over everything that happens before you touch y = x + 1. Dr. S And what does that do for you? KS It fixed the starting point to ( 1,1 instead of (0,0. Dr. S Do we know the new path has the right number of U and R steps? KS It worked in this example: Dr. S And how do we undo that? RG Reflect back over the line y = x + 1. MH So to go from D to D, take the inverse of the path up until its first intersection with y = x+1, then keep the remainder of the path the same. To go back, repeat the process again. Dr. S Why doesn t this work for a Catalan path? KS Because it would never meet the line y = x+1. And if you reflected the whole path, then the endpoint would no longer be at (n,n. Dr. S How does everyone feel about that?

17 Combinatorics 15 [C] Find a formula for the nth Catalan number C n, using your answers from (a and (b. SG When you count D you ll have, according to the lattice path counting formula from #49, ( ( 2n 2n or n 1 n + 1 MS So in total, C n = ( ( 2n 2n. n n + 1 Dr. S So what are we to make of the conjecture from before: C n = 1 2n n + 1( n? Might one or both results be wrong? This is probably more of an algebra problem now than a combinatorics problem. SG If we divide the left side by ( 2n n, we get ( 2n ( n 2n ( 2n n+1 n+1 ( 2n = 1 ( 2n n n and maybe we can simplify that quotient somehow. Dr. S Is there a way to handle the denominator in ( 2n (2n! n+1 = (n 1!(n + 1! MR But (n + 1! = n!(n + 1, since all you re doing is adding an extra factor at the This seems to be all end. MH And (n 1! = n! 1 n. KS And (n 1! is just included as part of n!. SG If you factor out ( 2n ( 2n n MH We got another way: ( 2n n n you have ( 2n n + 1 ( 2n n + 1 = (2n! n!n! (2n! (n 1!(n + 1! = (2n! ( n!n! 1 n!n! (n + 1!(n 1! ( 2n (1 n = n n + 1 ( 2n 1 = n n + 1 = (2n! n!n! (2n! n!( 1 n (n + 1n! = (2n! n!n! (2n! ( n n!n! n + 1 = (2n! ( n 1 n!n! n + 1 = ( 2n n We just factored last, instead of right away. 1 n + 1 you need to know about factorials to complete this argument.

18 16 Chapter 1. What is Combinatorics? Combinatorics October 24 27: Review Problems 1. How many different n-letter sequences can be made from the letters A, B, C, and D, if the sequence cannot include the word BAD in it? SY Our conjecture: the number of bad words that is, the words containing BAD is (n 2 4 n 3. Then the amount of good words: 4 n (n 2 4 n 3. QE D MR The 4 n describes the set of all n-letter words as a whole. SG Then (n 2 is the number of different options of where BAD is included in the word, and n 3 is the number of empty slots once BAD is included. SY For instance, the four-letter words including BAD are of two types: The five-letter words are of three types: xbad and BADx. BADxx xbadx xxbad. Dr. S About all the arithmetic in this expression: what does it mean combinatorially? MR The 4 n corresponds to a set of all functions, in this case functions {1,2,...,n} = N K = {A,B,C,D}. So every n-letter sequence corresponds to a function from N to K, of which there are K N = 4 n. SY n 2 is the number of choices for placing BAD in a sequence MR, MS Because the B needs to be at least three spaces over from the right end. You can t place the B in the last two spaces if you need to include BAD in the sequence. SY Then 4 n 3 is the number of choices raised to the number of x s. Since we took out the three spaces for B, A, and D, this is n 3. Dr. S Why are we multiplying (n 2 by 4 n 3? SY For each value of n, you have both a choice of where to put BAD, and then a choice of what letters to put in the free slots. Something like that. MS That feels right. 2. A fruit basket contains a different apples and b different bananas. How many ways are there for Jack to pick a piece of fruit, and then Jill to pick a piece of fruit? SY Jack has a + b choices, then Jill has a + b 1. MS Do a and b have to be greater than one? It seems like that s implied.

19 Combinatorics 17 MR Not necessarily; it s not stated in the problem. SG Jack has ( ( a+b 1 = a + b choices. Then Jill has a+b 1 1 = a + b 1 choices. MS Is it really important that the apples and bananas are different? If all the pieces of fruit are unique to begin with, why not just let C be the set of all fruit, and C = c = a + b. Then Jack has c choices, and Jill has c 1 choices, and we have a total of c(c 1. An expert observation. SG Or is it c + (c 1? Because there are c ways for Jack to choose, and (c 1 ways for Jill to choose. RG This is similar to the rollercoaster problem (#15. JT For a = 2 and b = 2 we got 12 possibilities. RG We got that, too. SY Then that s 4 3, agreeing with our answer: (2 + 2( = (4(3 = 12. MS So we can write A B = a + b = c, since every fruit in unique in a and b. Then (c }{{} (c 1 }{{} Jack s choices Jill s remaining choices SG And A B = A + B = a + b because A and B are disjoint sets. MR Because one set s elements are apples and the other s are bananas, and no fruit is both. So c = a + b and we have (a + b(a + b If all numbers from 1 to 10 n are listed, how many times is the digit 5 written? GU It seems like you pick up an additional 10% each time n increases by 1. RG But then you d eventually get to more than 100%, so it has to level off somewhere. SY This might be one of the trickier ones so far. If I were approaching this, I might find a way of counting all the numbers that don t include the digit 5. SG Then count the numbers that include the digit 5. Dr. S But each of them might have different number of 5s. What do the simple cases look like? MR For n = 3 I got 260. KS We got 300, and then 4000 for n = 4. In general, this pattern follows n 10 n 1. SG For example, in the n = 3 case you have

20 18 Chapter 1. What is Combinatorics? Combinatorics Numbers Five digits Total 200 MR If we presume that for n = 3 there are 300 fives, then each range of thousands which do not begin with 5 will have the digit times: Total 2700 But in the range where the numbers begin with 5, we have a five in the thousands place, and then 300 fives in the remaining places, for a total of =1300 So in total, there is = MH We understood where the 10 n 1 came from: when you re counting fives from 0 999, you end up counting the fives in the hundreds, not the thousands. SG That s true when you re counting the batches which don t start with a five, but not when it does. SY It s as though each time you add a digit (increasing n by 1, you end up multiplying by 10, and adding a power of 10: n = 1 n = 2 n = 3 n = Multiply by 10 and add 10. Then, multiply by 10 and add 100. Then, multiply by 10 and add Monday, November 7: #65 66

21 Combinatorics 19 say this graph is complete. 66. Draw five circles labeled Al, Sue, Don, Pam, and Jo. Find a way to draw red and green lines between people so that every pair of people is joined by a line and there is neither a triangle consisting entirely of red lines nor a triangle consisting entirely of green lines. MS I think we re starting with all the people in a pentagon shape. MR A Star of David should work. KS A pentagram, really. SY I thought that was supposed to be Satanic. KS We started by making all the outside lines green and the inside lines red. Dr. S Is there any other way to do it besides that? KS You have 10 lines total (I did = ( 5 2. Does it matter how many green lines and how many red? RG Does it matter if we arrange the dots differently? Dr. S Not as long as every pair of dots is still connected by a line. MR We may have found a different way. Dr. S Is there a connection between these? KS Each graph has the same number of red and green lines. MR And each person has two red and two green lines coming out. Dr. S But if you interchange Dan and Joe, and then Dan and Pam, you get the original graph again. Dr. S So are those two graphs different? Is there any advantage to considering these graphs to be the same or different? KS It s like a reflection. Dr. S In fact, we will define these two graphs to be isomorphic: identical but for a renaming of the vertices. 65. Prove that, in any group of six people, either three must be mutual friends, or three must be perfect strangers. (That is, the Ramsey number R(3,3 = 6.

22 20 Chapter 1. What is Combinatorics? Combinatorics SY Let s start by following the pentagram example above, using half the total number of edges of the complete graph. Dr. S How many edges total would there be in the complete graph K 6? SG Thirty (= 6 5 since each vertex must connect to the five others. JT I think that s too many. I counted fifteen. SG Because order doesn t matter when you connect an edge. Dr. S So we may want to try connecting close to half of that, because we re looking for a balance between having enough edges to eliminate groups of perfect strangers, but not so many edges that we create groups of mutual friends. Dr. S Here s a hint to get you started. Single out a vertex (say, vertex 1. If you look at 1 s friends (vertices to which it connects and strangers (vertices to which it does not, how many of each must there be? Is it possible for there to be less than three of each? Class No, because there are five other vertices to choose from. So 1 is either connected to, or disconnected from, at least three other vertices. Dr. S Now, without loss of generality, let s say that vertex 1 has three friends, 2, 4, and 6. What can we then say about vertex 2? SG Then, if we want to avoid making a triangle, 2 can connect only to 3 and 5. Dr. S Right now, what relationship exists between 2, 4, and 6 the friends of 1? KS They re friends of friends, but not friends. They re mutually disconnected. Dr. S So condition B (three mutual strangers is satisfied by 2, 4, and 6, unless two of them are friends. What happens then? SG Condition A (three mutual friends holds, since they are both friends with 1. Dr. S Is that enough? MH If there is an edge anywhere between 2, 4, and 6, then that edge will make a triangle with 1, giving three mutual friends.