3. Elementary Counting Problems 4.1,4.2. Binomial and Multinomial Theorems 2. Mathematical Induction Prof. Tesler Math 184A Fall 2017 Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 1 / 38 Multiplication rule Combinatorics is a branch of Mathematics that deals with systematic methods of counting things. Example How many outcomes (x, y, z) are possible, where x = roll of a 6-sided die; y = value of a coin flip; z = card drawn from a 52 card deck? (6 choices of x) (2 choices of y) (52 choices of z) = 624 × × Multiplication rule The number of sequences (x1, x2,..., xk) where there are n1 choices of x1, n2 choices of x2,..., nk choices of xk is n1 n2 nk. · ··· This assumes the number of choices of xi is a constant ni that doesn’t depend on the other choices. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 2 / 38 Cartesian product The Cartesian Product of sets A and B is A B = f (x, y): x A, y B g By the Multiplication× Rule, this has size2 jA 2Bj = jAj jBj. × · Example: f1, 2g f3, 4, 5g = f(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)g × The Cartesian product of sets A, B, and C is A B C = f (x, y, z): x A, y B, z C g This has size×jA ×B Cj = jAj jBj jC2j. 2 2 × × · · This extends to any number of sets. Example Roll of a 6-sided die A = f1, 2, 3, 4, 5, 6g jAj = 6 Value of a coin flip B = fH, Tg jBj = 2 Cards C = fA , 2 ,...g jCj = 52 ~ ~ The example on the previous slide becomes jA B Cj = 6 2 52 = 624. × × · · Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 3 / 38 Notation We often will need an n-element set. For n > 1, define [n] = f1, 2, 3,..., ng and also [0] = . ; Again, you may have seen [x] used for greatest integer, but we instead use x for floor and x for ceiling. b c d e Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 4 / 38 Powers Let A be a set. Ak = A A A (k times) × × · · · × jAkj = jAjk Example [2] = f1, 2g, with size j[2]j = j f1, 2g j = 2. [2]3 = f(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)g j[2]3j = 23 = 8 Example How many k letter strings are there over an n letter alphabet? 3-letter strings over the alphabet fa, b,..., zg: aaa, aab, aac,..., zzy, zzz There are 263 of them. In general, there are nk strings. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 5 / 38 Power set The power set of a set is the set of all of its subsets: P(S) = f A : A S g ⊆ P([3]) = P(f1, 2, 3g) = f , f1g , f2g , f3g , f1, 2g , f1, 3g , f2, 3g , f1, 2, 3gg ; Be careful on use of vs. : 2 ⊂ 1 < P([3]) P([3]) f1g P([3]) f1g , f2g , f1, 2g P([3]) ; 2 2 2 f g P([3]) ff1gg P([3]) ff1g , f2g , f1, 2gg P([3]) ; ⊂ ⊂ ⊂ How big is jP(S)j? Equivalently, how many subsets does a set have? Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 6 / 38 Number of subsets of an n-element set First solution How many subsets does an n element set have? We’ll use [n]; the solution will work for any set of size n, but it’s easier to work with a specific set. Make a sequence of decisions: Include 1 or not? Include 2 or not? Include··· n or not? Total: (2 choices)(2 choices) (2 choices) = 2n ··· Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 7 / 38 Number of subsets of an n-element set First solution Include 1? ; No Yes 1 Include 2? ; { } 2 1 1, 2 Include 3? ; { } { } { } ([3]) 3 2 2, 3 1 1, 3 1, 2 1, 2, 3 P ; { } { } { } { } { } { } { } Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 8 / 38 Number of subsets of an n-element set Second solution Consider a subset S [n]. ⊆ Form the word w1 wn or a sequence (w1,..., wn) ··· 1 if i S; wi = 2 0 otherwise. Example: The subset S = f1, 3, 4g of [5] is encoded as a word 10110 or as a sequence (1, 0, 1, 1, 0). Each subset of [n] gives a unique word in f0, 1gn and vice-versa. j f0, 1gn j = 2n, so there are 2n words and thus 2n subsets. This is called a bijective proof . Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 9 / 38 Function terminology Consider a function f : P Q. For element x in the set P, the function assigns a value f!(x) in the set Q. f is one-to-one iff for all x, y P, when f (x) = f (y) then x = y. 2 This is also called an injection. This means every element of Q either has exactly one inverse, or has no inverse. If f is one-to-one then jPj 6 jQj. f is onto iff for every z Q, there is at least one x P with f (x) = z. 2 2 This is also called a surjection. This means every element of Q has at least one inverse. If f is onto then jPj > jQj. f is a bijection iff it is one-to-one and onto. This means every element of Q has exactly one inverse. If f is a bijection then jPj = jQj. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 10 / 38 Function terminology f : P Q ! P Q For all functions, exactly one arrow leaves each element of P. One-to-one / Injection: At most one arrow hits each element of Q. Onto / Surjection: At least one arrow hits each element of Q. Bijection: Exactly one arrow hits each element of Q. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 11 / 38 Number of subsets of an n-element set Second solution, continued Define f : P([n]) f0, 1gn as follows: ! for S [n], form the word f (S) = w = w1 wn, where ⊆ ··· 1 if i S; wi = 2 0 otherwise. f is one-to-one: Suppose f (S) = f (T) = w. We need to show this requires S = T. Both S and T are subsets of [n]. For each i = 1,..., n, if wi = 1 then i S and i T, 2 2 while if wi = 0 then i < S and i < T. Thus, S and T have the exact same elements, so S = T. f is onto: Given w f0, 1gn, we must construct an inverse in the domain. There may2 be more than one inverse; we just have to construct one. S = f i [n]: wi = 1 g is in the domain and satisfies f (S) = w. 2 Thus, f is a bijection. So jP([n])j (the number of subsets of an n-element set) equals j f0, 1gn j = 2n. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 12 / 38 Number of subsets of an n-element set Third solution We will use Mathematical Induction (Chapter 2) to prove that the n number of subsets of [n] is 2 , for all integers n > 0: In general, the goal is to prove that a statement is true for all integers n > n0. Often, n0 is 0 or 1, but that’s not required. Base case: Show that the statement is true for n = n0. Sometimes it’s necessary to prove it specially for several other small values of n. Induction step: Assume that the statement holds for n. Use that to prove that it holds true for n + 1. Conclusion: the statement holds for all integers n > n0. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 13 / 38 Number of subsets of an n-element set Third solution n For all integers n > 0, the number of subsets of [n] is 2 . Base case First we show the statement is true for the smallest value of n (in this case, n = 0). When n = 0, [n] = [0] = has just one subset, which is . ; ; The formula gives 2n = 20 = 1. These agree, so the statement holds for the base case. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 14 / 38 Number of subsets of an n-element set Third solution Induction step n For some n > 0, assume that the number of subsets of [n] is 2 . Use that to prove the number of subsets of [n + 1] is 2n+1 Split the subsets of [n + 1] into P Q, where [ P is the set of subsets of [n + 1] that don’t have n + 1, and Q is the set of subsets of [n + 1] that do have n + 1. P is the set of subsets of [n]. By the induction hypothesis, jPj = 2n. Insert n + 1 into each set in P to form Q. Thus, jPj = jQj. E.g., for n = 2: P = , f1g , f2g , f1, 2g ; Q = f3g , f1, 3g , f2, 3g , f1, 2, 3g The total number of subsets of [n + 1] is jPj + jQj = 2(2n) = 2 n+1. (This is an example of the Addition Rule, to be covered next.) n Conclusion: For all integers n > 0, the number of subsets of [n] is 2 . Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 15 / 38 Addition rule Count the number of days in a year, as follows.