MATH 402 : 2017 page 1 of 6

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1 ADMINISTRATION What? Math 40: Enumerative Combinatorics Who? Me: Professor Greg Smith You: students interested in combinatorics When and Where? lectures: slot 00 office hours: Tuesdays at :30 6:30 and Wednesday 7:30 8:30 in 0 Jeffery Hall assessment: homework (40%: late problems sets are not accepted project (40%: detailed discussions in future lectures exam (0%: scheduled by registrar How? DO; one learns by working problems TALK mathematics with everyone you can READ; use multiple sources Why? problem solving skills communication skills beauty and elegance OVERVIEW What is enumerative combinatorics? In enumerative combinatorics, one counts the number of elements in a finite set or, more often, simultaneously count the number of elements in an infinite collection of sets. Subsets, partitions, and permutations of {,,...,n} are classic examples. Although these topics play a fundemental role in most areas of mathematics and computer science, the actual prerequisites are surprisingly small. I hope that this course will change the way you think about equations and power series. By design, the course will emphasis influential examples, techniques, and communication. Notation. Throughout the course, N := {0,,, 3,...} denotes the set of nonnegative integers. THE PIGEONHOLE PRINCIPLE This seemingly obvious statement produces unexpected results. Theorem (Pigeonhole Principle. If n objects are placed in r boxes where r < n, then at least one of the boxes contains more than one object. Proof. To establish a contradiction, assume that no box has more than one object. If there is no box with at least two objects, then each of the r boxes has either zero or one object in it. Let m be the number of boxes with zero objects in them; certainly m 0. It follows that there are r m boxes with one object. However, this means that the total number of objects placed in the r boxes is MATH 40 : 07 page of 6

2 r m r < n; contradiction. Therefore, our assumption is false and there exists at least one box with more than one object. Problem. Show that, if you pick cards from a standard deck of cards, then at least will be of the same suit. Solution. Each of the cards can belong to one of 4 suits. By the Pigeonhole Principle, two or more must belong to the same suit. Problem. Consider the sequence,3,7,,3,...; the jth element is a j := j+ for j N. If n is any odd integer, then this sequence contains an element that is divisible by n. Every odd integer has a multiple that is one less than a power of two. Solution. Focus on the first n elements of the sequence. If one of them is divisible by n, then we are done. Otherwise, consider their remainders modulo n; write a j = q j n + r j where 0 < r j < n and q j := a j /n. Since there are n distinct remainders, the Pigeonhole Principle implies that two elements of the sequence have the same remainder. If r j = r k where n > j > k > 0, then we have (q j q k n = a j a k = ( j+ ( k+ = j+ k+ = k+ ( j k = k+ a j k. It follows that n divides a j k, because k+ and n are relatively prime. Problem*. Given 0 points in the unit square, there exists points that are at most /3 units apart, and there exists 3 contained in a circle of radius /. Solution. Partition the unit square into nine squares of equal area. Since we have 0 points and 9 smaller squares, the Pigeonhole Principle implies that points lie in the same square. In a square, the points on the opposite corners have greatest distance between them. In our case, the Pythagorean Theorem implies that the distance is at most /3. Proposition* (From The Book. Given n integers a,a,...,a n, there is always a set of consecutive numbers a j+,a j+,...,a k with j < k whose sum k i= j+ a i is a multiple of n. Proof. Set N := {0,a,a +a,...,a +a + +a n } and R := {0,,...,n }. Consider the map f : N R where f (m is the remainder of m upon divison by n. Since N = n + > n = R, the Pigeonhole Principle implies that there are sums a + a + + a j and a + a + + a k with 0 j < k n and the same remainder. It follows that is a multiple of n. a j+ + a j+ + + a k = (a + a + + a j (a + a + + a k MATH 40 : 07 page of 6

3 THE GENERALIZED PIGEONHOLE PRINCIPLE Paul Erdős liked to talk about the The Book, in which God maintains the perfect proofs for mathematical theorems, following the dictum of G. H. Hardy that there is no premanent place for ugly mathematics. Erdős also said that you need not believe in God but, as a mathematician, you should believe in The Book. Notation. For n N, we set [n] := {,,3,...,n}, so [0] =, [] = {}, and [] = {,}. Proposition (From The Book. In any subset of [n] of cardinality n +, there are two relatively prime integers. Proof. Partition [n] into the n disjoint subsets (boxes: {,},{3,4},{,6},...,{n,n}. View the given subset as placing n + objects into these boxes. By the Pigeonhole Principle, there is box with at least two objects. In other words, the subset contains two consecutive integers and two numbers that differ by are relatively prime. Proposition (From The Book. In any subset of [n] of cardinality n +, there exists a pair of integers such that one divides the other. Remark. When n = 4, what happens? The possible pairs such that one divides the other are (,,(,4,(,6,(,8,(3,6,(4,8, so... Proof. Write every number in the given subset in the form k m where m is an odd number between and n. Since there are n + numbers in the subset, but only n different odd parts, the Pigeonhole Principle shows that there must be two numbers with the same odd part. Hence, one is a multiple of the other. Remark. Both Propositions are false if one replaces n + by n. {,4,6,...,n} and {n +,n +,...,n}. Indeed, consider the sets Theorem (Generalized Pigeonhole Principle. Let n and r be positive integers such that n > r. If n objects are placed in r boxes, then at least one of the boxes contains at least n r objects. Proof. In the language of mappings, the principle asserts that, for two finite sets N and R with N = n > r = R and any function f : N R, there exists an element m R with f (m nr. Otherwise, we would have f (m < n r for all a R and n = m R f (m ( < r nr = n which is absurd. Problem. Show that there are at least 4 people currently living in the Toronto Area who were born in the same hour of the same day of the same year. Solution. The verified oldest person lived less that 3 years (Jeanne Calment lived years, 64 days; , so we assume that all the residents of the Toronto Area are at most years old. As each year consists of at most 366 days, we have ((366(4 = boxes. Since there are at least 3794 Torontonians, the Generalized Pigeonhole Principle establishes that at least 3794/ > 4 people where born within 60 minutes of each other. Dirichlet used the Pigeonhole Principle to prove the existence of good rational approximations to irrational numbers. By taking the denominator to be q = n, it is easy to get an approximation whose error is at most /n. MATH 40 : 07 page 3 of 6

4 Theorem (Dirichlet 879. For any real number x and any positive integer n, there is a rational number p/q such that q n and x p q < nq q. Proof. Let {x} denote the fractional part of the real number x, so {x} := x x and 0 {x} <. Consider the n + numbers { jx}, where j n + and j Z, and the n boxes given by the intervals [0,/n,[/n,/n,...,[(n /n,n/n which partition [0,. By the Pigeonhole Principle, there exists an interval containing more than one of the numbers. If the numbers { jx} and {kx} with j > k belong to the same interval, then they differ by less than /n. Setting q := j k and p := jx kx, we have n > { jx} {kx} = jx jx kx + kx = ( j kx ( jx kx = qx p which yields x p/q < /nq. The Thue Siegel Roth theorem shows that this is essentially the tightest possible bound; the bound on rational approximation of algebraic numbers cannot be improved by increasing the exponent beyond. The next application of the Pigeonhole Principle is not immediate. Theorem* (Erdős Szekeres 939. In a sequence a,a,...,a mn+ of mn + distinct real numbers, there exists an increasing subsequence a i < a i < < a im+ where i < i < < i m+ of length m + or a deceasing subsequence a j > a j > > a jn+ with j < j < < j n+ of length n +. Proof. Associate to each element a i the length l i of a longest increasing subsequence starting at a i. If l i m+ for some i, then there is an increasing subsequence of length m+. Suppose otherwise, so we have l i m for i mn +. For the function a i l i mapping {a,a,...,a mn+ } to {,,...,m}, the Generalized Pigeonhole Principle implies that there exists some r [m] such that a i r for mn m + = n + elements. Let a j,a j,...,a jn+ with j < j < < j n+ be these elements. Look at two consecutive numbers a jk,a jk+ for some k n. If a jk < a jk+, then we could obtain an increasing subsequence of length r starting at a jk+ and consequently an increasing subsequence of length r + starting at a jk which cannot be since a jk r. Therefore, we obtain a decreasing subsequence a j > a j > > a jn+ of length n +. MATH 40 : 07 page 4 of 6

5 FIBONACCI NUMBERS Fibonacci considers the growth of an idealized (biologically unrealistic rabbit population, assuming that: a newly born pair of rabbits, one male, one female, are put in a field; rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits; rabbits never die and a mating pair always produces one new pair (one male, one female every month from the second month on. The puzzle that Fibonacci posed was: how many pairs will there be in one year? Definition. The Fibonacci numbers are defined by F 0 = 0, F =, and := + for all n. The first few numbers in the sequence are 0,,,,3,,8,3,,34,,89,44,33. This sequence has many remarkable properties. Problem. Show that F 0 + F + F + + = +. Solution. We proceed by induction on n. For n = 0, we have F 0 = 0 = = F, which establishes the base case. For n > 0, the induction hypothesis and the Fibonacci recurrence give F 0 + F + F + + = (F 0 + F + F = (+ + = (+ + = +. Problem. Prove that the Fibonacci number is even if and only n is divisible by 3. Solution. We proceed by induction. Since F 0 = 0, F =, and F =, the base case follows. For j > 0, suppose that F 3 j is even, F 3 j+ is odd, and F 3 j+ is odd. The next three Fibonacci numbers are which completes the induction. F 3( j+ = F 3 j+ + F 3 j+ = odd + odd = even F 3( j++ = F 3( j+ + F 3 j+ = even + odd = odd F 3( j++ = F 3( j++ + F 3( j+ = odd + even = odd, Theorem (Closed Formula. The Fibonacci numbers are given by = Proof. We can verify this formula as follows: F 0 = ( + 0 ( 0 = = 0 F = + = = = = = ( = ( + + = n ( n ( + + n ( n ( + + ( n ( + n ( 3+ ( n ( 3 n ( 6+ 4 n (! ( n ( 6 4 n ( n ( n n. = Fn. By mimicking the methods used to solve linear differential equations with constant coefficients, we can also rediscover this formula. MATH 40 : 07 page of 6

6 Proof. Suppose the recurrence relations = 0 has a solution of the form = x n. It follows that 0 = x n x n x n = x n (x x, so x = ±, 0. Since the Fibonacci recurrence ( is linear, the general solution has the form c + n ( + n c for some constants c and c. ( The initial conditions give 0 = c + c and = c + ( + ( c, so we obtain c = c and ( = c + ( ( = c + + = c. Therefore, the solution to the initial value problem is = ( + n ( n. Remark. This method generalizes to all linear recurrences. We will strength the analogy between recurrence relations and calculus later in the course. Definition. The Golden ratio is Some twentieth-century artists and architects have proportioned their works to approximate the golden ratio especially in the form of the golden rectangle, in which the ratio of the longer side to the shorter is the golden ratio believing this proportion to be aesthetically pleasing. The golden ratio also appears in some patterns in nature, including the spiral arrangement of leaves and other plant parts. + Problem*. Show that lim = + n. Solution. Set x n := + for n. Because > 0 for n > 0, we see that + = + > and x n > for n >. If the limit exists, then we have lim x n = x and n x = lim n x n = lim n + = lim n + + = lim = lim + n n + = lim n + = + lim = + n x n x = + + / + /x = + x x +. Hence, we obtain x x = 0. Since x, we conclude that x = + It remains to show that the limit exists. The equation x n = + ( ( implies that x n x n = + x n + x n = x n x n. = + = + = + x n x n x n. When x n x n < 0, it follows that x n x n < 0 and x n x n < x n x n x n x n = x n x n, so x n > x n > x n. Similarly, when x n x n > 0, we obtain x n < x n < x n. Since x =, x =, and x 3 = 3/, we have x > x 4 > x 6 > > x j > and x < x 3 < x < < x j < x j < x =. Hence, the odd term are increasing and bounded above, and the even terms are decreasing and bounded below, so these subsequences converge. The first paragraph show that the subsequences have the same limit, so the original sequence converges. MATH 40 : 07 page 6 of 6