2030 LECTURES. R. Craigen. Inclusion/Exclusion and Relations

Transcription

1 2030 LECTURES R. Craigen Inclusion/Exclusion and Relations

2 The Principle of Inclusion-Exclusion 7 ROS enumerates the union of disjoint sets. What if sets overlap? Some 17 out of 30 students in a class watch Game of Thrones, 20 students watch Big Bang Theory and 12 students watch both shows. How many students watch at least one of them? ROS says, for disjoint sets A, B, that A B = A + B. We might articulate a generalized ROS that works without the assumption of disjointness as follows: For any sets A, B, A B = A + B A B A glance at a Venn diagram shows why this rule is valid. The corresponding rule for 3 sets A, B, C which might overlap? A B C = A + B + C A B A C B C + A B C (Draw a Venn Diagram) What is the general rule?

3 Setting up Inclusion-Exclusion Note: My notation differs slightly from that of the text! Let S be a set of (say) N objects. Given conditions c 1, c 2,..., c r on elements of S, write N(c i ) for the number of elements of S satisfying c i. c i negates c i (i.e, c i says c i does not hold ). Thus, N(c i ) + N(c i ) = N. EG: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Then N = S = 10. Define: c 1 : element is odd c 2 : element is a prime c 3 : element is a power of 2 Then N(c 1 ) = 5; N(c 2 ) = 4; N(c) = 4 N(c 1 ) = 5; N(c 2 ) = 6 Now for conditions a and b write ab for both a and b hold. Similarly abc means all three of a, b and c hold

4 Enumerating sets by conditions Again with S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and conditions c 1 : odd; c 2 : prime; c 3 : power of 2, find: N(c 1 c 2 ) = 3; N(c 1 c 2 ) = 2; N(c 1 c 2 c 3 ) = 0 Having a condition on elements amounts to the same thing as having a subset of the original set. Condition c corresponds to the subset T of elements satisfying c, and N(c) = T. Restate our formulas for enumerating the unions of 2 and 3 subsets of some S, S = N, corresponding to conditions c 1, c 2 (and c 3 ): N(c 1 or c 2 ) = N(c 1 ) + N(c 2 ) N(c 1 c 2 ) N(c 1 or c 2 or c 3 ) = N(c 1 )+N(c 2 )+N(c 3 ) (N(c 1 c 2 )+N(c 1 c 3 )+N(c 2 c 3 ))+N(c 1 c 2 c 3 ) Write N 1 for the sum of numbers N(c i ), N 2 for the sum of numbers N(c i c j ) over all pairs with i j, and so on, so with 3 conditions N 2 = N(c 1 c 2 ) + N(c 2 c 3 ) + N(c 1 c 3 ), and N 3 = N(c 1 c 2 c 3 ). The above formulas reduce to N(c 1 or c 2 ) = N 1 N 2 and N(c 1 or c 2 or c 3 ) = N 1 N 2 + N 3

5 How many elements satisfy none of the conditions? More often we want to know how many elements of S satisfy none of the conditions. That is, enumerate the complement of a union of sets N(c 1 or c 2 or... or c r ) = N(c 1 c 2 c r ) Let us also write N 0 = N and N = N(c 1 c 2 c r ). For 2 conditions: N = N(c 1 c 2 ) = N N(c 1 or c 2 ) = N 0 (N 1 N 2 ) = N 0 N 1 +N 2 For 3 sets/conditions our formula is N = N(c 1 c 2 c 3 ) = N (N 1 N 2 + N 3 ) = N 0 N 1 + N 2 N 3 EG How many permutations of ABCDE do not contain the strings AB, BC or DE? Theorem 7.1 (Principle of Inclusion-Exclusion): ( Proof: ) N = N 0 N 1 + N 2 + ( 1) r N r

6 Examples of Inclusion-Exclusion EG: How many integers among 0, 1, 2,..., use all of the digits 1, 3, 4 and 7? (EG doesn t use 7; and use all 4) EG: How many solutions are there to the equation x 1 + x 2 + x 3 + x 4 = 20 in nonnegative integers x 1, x 2, x 3, x 4 where x 1 5, x 2 8, x 3 10, and x 4 13? EG: How many positive integers < 100 are divisible by neither 2 nor 5?

7 Counting relatively prime numbers Integers m, n are relatively prime if their greatest common divisor is 1 EG: Which pairs of numbers 6, 7, 8, 9, 10 are relatively prime? It is equivalent to say that m, n share no common prime divisor. If m n, the number of multiples of n up to m is m n. More generally this numbers is m n. EG: How many numbers up to 100 are divisible by 7? Relatively prime to 7? EG: How many numbers up to 1000 are relatively prime to 30?

8 Euler Phi function How many numbers n are relatively prime to n? For positive integer n = p e 1 1 pe 2 2 per r = r i=1 p e i i. The Euler φ function of n, φ(n) (Also known as the totient function) is defined as the number of positive integers n which are relatively prime to n. EG: φ(10) = 4, φ(11) = 10. Theorem 7.2: If n = φ(n) = n r i=1 p e i i, then r ) (1 1pi i=1 ) ) = n (1 1p1 (1 1pr

9 Derangements A place for everything and nothing in its place A derangement of a set is a permutation of that set (from a given initial order) in which no element is returned to its initial position. EG: How many derangements of {1, 2, 3, 4}? The number of derangements of an n-set is denoted D n So D 1 = 0; D 2 = 1; D 3 = 2; D 4 = 9 Formula for D n : D n = n! [1 ] ( 1)n ! 12! 13! n! = n! n ( 1) i n! Check above numbers. Evaluate D 5 = 44.

10 The probability of a derangement EG: n gentlemen check their hats as they arrive at a party. After much drinking they all leave, in their drunken state randomly taking hats. What is the probability that none of the gentlemen receive their own hat back? EG: n students write a quiz. The papers are collected, shuffled, and returned randomly to the students for marking. What is the probability that no student receives their own paper to mark? The probability we arrive at is p n = Dn n! = n first few values of p n : ( 1) i i!. Here are the n p n What is familiar about this formula? What do you notice about its values as n grows? Do you think this converges? To what value? How close is it in general? Could we use this observation to find D n for large n?

11 Counting onto functions Distinguishable balls into distinguishable bins (none empty) How many functions from an n-set A onto a k-set B? What is the answer if we don t require functions be onto? k n functions. S: set of functions f : A B = {b 1, b 2,..., b k } c 1 : b 1 is not in the range of f c 2,..., c k : b 2,..., b k respectively not in the range of f The required number is = k n k(k 1) n + N = N 0 N ( 1) k N k ( ) ( ) k k (k 2) n + ( 1) k (k k) n 2 k k 1 ( ) k k 1 = ( 1) i (k i) n ( 1) i or, k! (k i)n i i!(k i)!

12 Formula for Stirling numbers of the second kind: S(n, k) Distinguishable balls into indistinguishable bins (none empty) Another way to count distributions of n distinguishable balls into k distinguishable bins: Task 1: Put the balls into k indistinguishable bins Task 2: Put k labels on the containers (making them distinguishable) Task 1 can be done S(n, k) ways (by definition). No bin is empty and balls distinct so bins are distinguished by their contents. So Task 2 permute k labels. By ROP We have It follows that k 1 ( k ( 1) i i ) (k i) n = k!s(n, k) S(n, k) = 1 k 1 ( k ( 1) i k! i ) (k i) n

13 A couple of examples EG: How many onto functions from a 6-set to a 4-set? (n = 6 and k = 4). Ans: 4 1 ( ) 4 ( 1) i (4 i) 6 = = 4( ) = 1560 i EG: How many onto functions from a 4-set to a 6-set? (n = 4 and k = 6). Ans: 0 EG: How many ways to distribute 6 distinguished balls into 4 indistinguished bins leaving none empty? (n = 6 and k = 4). Ans: S(n, k) = ( ) 4 ( 1) i (4 i) 6 = ! i 24 = 65 - In each case, what if n = k? - The sum in both formulas goes to i = k 1. Why not to k?

14 Counting how many objects satisfy exactly m conditions Theorem 7.4: The number of objects satisfying exactly m out of r conditions, where m r, is ( ) ( ) ( ) m + 1 m + 2 m + r m N m N m+1 + N m+2 +( 1) r m 1 2 r m r m = ( 1) i ( m + i i ) N m+i Lemma (for proving Theorem 7.4): For all suitable m, i, j: ( )( ) ( )( ) m + j m + i m + j j = m + i i m i N r EG: How many primes up to 100? How many are products of two distinct primes? EG: Generalized hatcheck problem.