Lecture 21 Principle of Inclusion and Exclusion

Transcription

1 Lecture 21 Principle of Inclusion and Exclusion Holden Lee and Yoni Miller 5/6/11 1 Introduction and first exaples We start off with an exaple Exaple 11: At Sunnydale High School there are 28 students in algebra class, 30 students in biology class, and 8 students in both classes How any students are in either algebra or biology class? Solution Let A denote the set of students in algebra class and B denote the set of students in biology class To find the nuber of students in either class, we first add up the students in each class: A + B However, this counts the students in both classes twice Thus we have to subtract the once: A B This shows A B = A + B A B A B = = 50, so there are 50 students in at least one of the two classes The sae reasoning wors with three sets Exaple 12: At Sunnydale High School there are 55 students in either algebra, biology, or cheistry class 28 students in algebra class 30 students in biology class 24 students in cheistry class 8 students in both algebra and biology

2 16 students in both biology and cheistry 5 students in both algebra and cheistry How any students are in all three classes? Solution Let A, B, C denote the set of students in algebra, biology, and cheistry class, respectively Then A B C is the set of students in one of the three classes, A B is the set of students in both algebra and biology, and so forth To count the nuber of students in all three classes, ie count A B C, we can first add all the nuber of students in all three classes: A + B + C However, now we ve counted the students in two classes too any ties So we subtract out the students who are in each pair of classes: A B B C A C For students who are in two classes, we ve counted the twice, then subtracted the once, so they re counted once But for students in all three classes, we counted the 3 ties, then subtracted the 3 ties Thus we need to add the again: Thus + A B C A B C = A + B + C A B B C A C + A B C 55 = A B C 2

3 Thus A B C = 2, ie there are 2 students in all three classes The sae reasoning wors with an arbitrary nuber of sets; we state the general result in the following theore Theore 13 (Principle of Inclusion and Exclusion (PIE)): Let A 1,, A n be sets Define S = 1 i 1 < <i n A i 1 A i, ie S is a su over all choices of subsets For = 1, it s siply A 1 + A 2 + A A and for = n, it represents one su, Then A 1 A 2 A 3 A n A 1 A n = S 1 S 2 + S ( 1) n 1 S n Proof Consider an eleent x A 1 A n It is counted once on the left hand side, so we need to show that it is counted once on the right hand side as well Suppose x is in the sets A j1,, A j, but not in the rest of the sets There are ( ) sets {i1,, i } such that x A i1 A i, since {i 1,, i } has to be a subset of {j 1,, j } Thus x is counted ( ) ties in the su 1 i 1 < <i n A i 1 A i, and hence is counted ( ) ( ) ( ) + + ( 1) ) ( + ( 1) ) = on the right hand side By the Binoial Theore, 1 ( ) ( (1 1), so ( ) ( ) ( ) + + ( 1) 1 = 1, 1 2 showing x is counted once in the RHS If x A 1 A n, clearly x is not counted in either the LHS or the RHS Since each eleent is counted the sae nuber of ties on either side, the theore follows Exaple 14: Find the nuber of positive integers less than or equal to 1000 that are divisible by 7, 10, or 15 Solution For a positive integer, let A denote the set of integers in {1, 2,, 1000} that are divisible by We want to find A 7 A 10 A 15 Note that 1000 A = where y denotes the greatest integer less than or equal to y Indeed, the ultiples of that are less than 1000 are exactly, 2,, 1000 Note also that A A l = A lc(,l) since a nuber is divisible by both and l if and only if it is divisible by lc(, l) Using PIE, we get A 7 A 10 A 15 = A 7 + A 10 + A 15 A 7 A 10 A 7 A 15 A 10 A 15 + A 7 A 10 A 15 = A 7 + A 10 + A 15 A 70 A 105 A 30 + A = = =

4 Suppose instead we want to find the su of positive integers satisfying the above properties, not the nuber of positive integers What do we do? We replace each ter S in the su above by x S x, so, rather than contributing 1 to the value S, x contributes the value x to S This is an exaple of the following ore generalized version of inclusion-exclusion Theore 15: Let f : A 1 A n R be a function Then x A 1 A n f(x) = n =1 1 i 1 < <i n = 1 i 1 n ( 1) x A i1 f(x) x A i1 A i f(x) 1 i 1 <i 2 n x A i1 A i2 f(x) + + ( 1) n 1 Note that taing f to be identically 1, we get the original stateent of PIE x A 1 A n f(x) Proof The proof is the sae as above, except now an eleent x contained in of the sets contributes f(x) to the left hand side and (( ) ( 1 ) ( ( 1) +1 )) f(x) to the right hand side 2 Perutations and surjections Exaple 21: As n people wal into a dinner party, they reove their hats When they leave, the hats are randoly returned to the, so that all n! atchings between people and hats are equally liely What is the probability that nobody gets his or her own hat bac? Solution Labeling the people and the hats 1, 2,, n, a atching between people and hats corresponds to a perutation π on 1, 2,, n, where π(i) = j if person i receives person j s hat So an equivalent forulation is the following: Given a rando perutation on {1, 2,, n} (that is, a one-to-one function on {1, 2,, n}), what is the probability that there is no fixed point, ie no i such that π(i) = i? Let A be the set of perutations having a fixed point, and D be the set of perutations having no fixed point (these perutations are called derangeents) We need to find A To use PIE, we want to express A as a union of sets whose sizes are easy to calculate Note that π having a fixed point eans either 1 is a fixed point, or 2 is a fixed point, or 3 is a fixed point, or so on Hence A = A 1 A n where A i is the set of perutations π such that i is a fixed point, ie π(i) = i By PIE, we have n ( 1) 1 A i1 A i (1) =1 1 i 1 <<i n An eleent π of A i1 A i ust have π(i 1 ) = i 1,, π(i ) = i, but can perute the other n eleents in any way Hence A i1 A i = (n )! 4

5 Since there are ( n ) sets of size, (1) becoes n ( 1) 1 (n )! = =1 D = n! Thus the desired probability is n =0 ( 1) n!! n ( 1) 1! n! =1 D n! = n ( 1)! =0 Readers failiar with calculus ay recognize that truncated series for 1 appears above: e We calculated that Hence D n! e = n! 1 e = 1 1 1! + 1 2! 1 3! + 1 4! D = n! (1 =n+1 ( 1)! ) ( 1)n + 11! 12! n! ( n! ) ( (n + 1)!, n! 1 (n + 1)! 2, 1 ) 2 Hence D equals n! rounded to the nearest integer, ie D = n! + 1 e e 2 and the probability equals 1 p = e + 1 2n! In particular, as the nuber of hats n goes to infinity, the probability that all people get their hats bac approaches 1 e Note: What aes this proof so fascinating, is that it shows us whether it s five people who derange their hats or a thousand or billion, it will generally be the sae probability Exaple 22: Suppose n Find the nuber of surjective functions f : {1, 2,, } {1, 2,, n}, ie functions f such that f(x) = has a solution for any {1, 2,, n} Solution Let S be the set of surjective functions As in the previous exaple, we find it ore convenient to consider the copleent of the desired set, then subtract it fro the set of all functions So let A = S c be the set of non-surjective functions We can express A as a union A = A 1 A n where A is the set of functions f : {1, 2,, } {1, 2,, n} whose range does not include Then by PIE, n ( 1) +1 A i1 A i (2) =1 1 i 1 < <i n 5

6 Now A i1 A i consists of all functions that iss the values i 1,, i This leaves n possibilities for each of f(1),, f(), so there are (n ) such functions Since there are ( n ) subsets of size, (2) becoes n ( 1) 1 (n ) Since there are n functions fro {1,, } to {1,, n}, we get S = n 3 Probles =1 =1 n ( 1) 1 (n ) = n ( 1) (n ) 1 (AHSME 1983/26) The probability that event A occurs is 3 ; the probability that 4 event B occurs is 2 Let p be the probability that both A and B occur Find the 3 sallest interval necessarily containing p 2 At Sunnydale High School there are =0 44 students in either algebra, biology, or cheistry class 25 students in biology class 23 students in cheistry class 13 students in both algebra and biology 9 students in both biology and cheistry 10 students in both algebra and cheistry 6 students in all three classes How any students are in algebra class? 3 Four couples are sitting in a row Find the nuber of arrangeents in which no person is sitting next to his or her partner What if they are sitting in a circle? 4 Find the nuber of five-digit cobinations fro the set {1, 2, 3, 4, 5} in which: (a) Soe digit appears at least three ties (b) No digit appears ore than twice 5 Let {p 1,, p r } be a set of pries Show that the nuber of positive integers less than or equal to n and relatively prie to p 1,, p r is equal to ) ) x (1 1p1 (1 1pr + 2 r θ for soe 1 < θ < 1 6 Find the su of all positive integers less than or equal to 1000 that are not divisible by 7, 10, or 15 6

7 7 Find the nuber of solutions to x 1 + x 2 + x 3 + x 4 + x 5 + x 6 = 15 such that each x i is an integer and 0 x i 4 8 Specially ared boxes of ChocoBalls cereal contain one of four puzzle pieces, each with probability 1 Once you have all four puzzle pieces, you can ail the in and 4 receive a coupon for ore ChocoBalls cereal (a) If Ji buys n specially ared boxes of ChocoBalls cereal, what is the probability that he gets all four puzzle pieces? (b) What is the expected nuber of boxes that Ji has to buy to get all four puzzle pieces? 9 Find the nuber of perutations π of {1, 2,, n} such that π(i + 1) π(i) + 1 for all 1 i n 10 Let a n be the nuber of perutations π of {1, 2,, 2n} such that π(i+1) π(i) n a for each 1 i 2n 1 Find a n and show that li n n = 1 (2n)! e 11 (TST 2004/2) Assue n is a positive integer Consider sequences a 0, a 1,, a n such that a i {1, 2,, n} for each i and a n = a 0 (a) Call such a sequence good if for all i = 1, 2,, n, a i a i 1 i (od n) Suppose n is odd Find the nuber of good sequences (b) Call such a sequence great if for all i = 1, 2,, n, a i a i 1 i, 2i (od n) Suppose that n is an odd prie Find the nuber of great sequences (Note: We solved this proble using generating functions and coplex nubers in lecture 11 Try to do it with PIE this tie) 12 (IMO 1991/3) Let S = {1, 2, 3,, 280} Find the inial natural nuber n such that in any n-eleent subset of S there are five nubers that are pairwise relatively prie 13 (USAMO 1994/5) Let U, σ(u), and π(u) denote the nuber of eleents, the su, and the product, respectively, of a finite set U of positive integers (If U is the epty set, U = 0, σ(u) = 0, and π(u) = 1) Let S be a finite set of positive integers As usual, let ( ) n denote n! Prove that!(n )! ( ) σ(u) ( 1) U = π(s) S for all integers σ(s) References U S [1] The Inclusion-Exclusion Principle, Math232/Inclusion-Exclusionpdf 7