Inclusion-Exclusion Rosen 6.5 & 6.6 Longin Jan Latecki basd on Slides by Max Welling, University of California, Irvine Vardges Melkonian, Ohio University, Athens

6.5 Inclusion-Exclusion U A B It’s simply a matter of not over-counting the blue area in the intersection.

Example on Inclusion/Exclusion Rule (2 sets) Question: How many integers from 1 through 100 are multiples of 3 or multiples of 7 ? Solution: Let A=the set of integers from 1 through 100 which are multiples of 3; B = the set of integers from 1 through 100 which are multiples of 7. Then we want to find n(A  B). First note that A  B is the set of integers from 1 through 100 which are multiples of 21 . n(A  B) = n(A) + n(B) - n(A  B) (by incl./excl. rule) = 33 + 14 – 4 = 43 (by counting the elements of the three lists)

Now three Sets area = 4-3=1 U C area = 2-1=1 area = 1 B A Image a blue circle has area 4. The intersections between 2 circles have area 2 and the intersection between three circles 1. What is the total area covered? A=4+4+4 – 2 -2 -2 + 1 = 12 – 6 + 1 = 7.

Example on Inclusion/Exclusion Rule (3 sets) 3 headache drugs – A,B, and C – were tested on 40 subjects. The results of tests: 23 reported relief from drug A; 18 reported relief from drug B; 31 reported relief from drug C; 11 reported relief from both drugs A and B; 19 reported relief from both drugs A and C; 14 reported relief from both drugs B and C; 37 reported relief from at least one of the drugs. Questions: 1) How many people got relief from none of the drugs? 2) How many people got relief from all 3 drugs? 3) How many people got relief from A only?

Example on Inclusion/Exclusion Rule (3 sets) We are given: n(A)=23, n(B)=18, n(C)=31, n(A  B)=11, n(A  C)=19, n(B  C)=14 , n(S)=40, n(A  B  C)=37 Q1) How many people got relief from none of the drugs? By difference rule, n((A  B  C)c ) = n(S) – n(A  B  C) = 40 - 37 = 3 S A B C

Example on Inclusion/Exclusion Rule (3 sets) Q2) How many people got relief from all 3 drugs? By inclusion/exclusion rule: n(A  B  C) = n(A  B  C) - n(A) - n(B) - n(C) + n(A  B) + n(A  C) + n(B  C) = 37 – 23 – 18 – 31 + 11 + 19 + 14 = 9 Q3) How many people got relief from A only? n(A – (B  C)) (by inclusion/exclusion rule) = n(A) – n(A  B) - n(A  C) + n(A  B  C) = 23 – 11 – 19 + 9 = 2

The Principle of Inclusion-Exclusion Proof: We show that each element is counted exactly once. Assume element ‘a’ is in r sets out of the n sets A1,...,An. -The first term counts ‘a’ r-times=C(r,1). -The second term counts ‘a’ -C(r,2) times (there are C(r,2) pairs in a set of r elements). -The k’th term counts ‘a’ -C(r,k) times (there are C(r,k) k-subsets in a set of r elements). -... - If k=r then there are precisely (-1)^(r+1) C(r,r) terms. - For k>r ‘a’ is not in the intersection: it is counted 0 times. Total: C(r,1)-C(r,2)+...+(-1)^(r+1)C(r,r) Now use: to show that each element is counted exactly once.

Applications of Incl.-Excl. We can use inclusion/exclusion to count the number of members of a set that do not have a bunch of properties: P1,P2,...,Pn. Call N(Pi,Pj,Pk,...) the number of elements of a set that do have properties Pi, Pj, Pk,.... and N the total number of elements in the set. By inclusion/exclusion we then have: Theorem: Let Ai be the subset of elements of a set A that has property Pi. The number of elements in a set A that do not have properties P1,...Pn is given then by:

A Picture U C B A

Examples Compute the number of solutions to x1+x2+x3=11 where x1,x2,x3 non-negative integers and x1 <=3, x2<=4, x4<=6. P1: x1 > 3 P2: x2 > 4 P3: x3 > 6 The solution must have non of the properties P1,P2,P3. The solution of a problem x1+x2+x3=11 with constraints x1 > 3 is solved as follows: 7 more balls 4 balls in basket x1 already. Total number of ways: C(7+3-1,7)=36 x1 x2 x3 Therefore: N-N(P1)-N(P2)-N(P3)+N(P1,P2)+N(P2,P3)+N(P1,P3)-N(P1,P2,P3) = C(11+3-1,11) – C(7+3-1,7) – C(6+3-1,6) – C(4+3-1,4) + … – 0.

Connection with De Morgan’s law So we have 2 ways to solve the last example: x1+x2+x3 = 11 such that non of the following properties hold: P1: x1 > 3 P2: x2 > 4 P3: x3 > 6 or x1+x2+x3=11 such all of the following properties hold: Q1=NOT P1: 0<x1<=3 Q2=NOT P2: 0<x2<=4 Q3=NOT P3: 0<x3<=6 Sometimes this is easier to compute.

Number of Onto-Functions Onto or surjective functions: A function f from A to B is onto if for every element b in B there is an element a in A with f(a)=b. If we have m elements in A and n in B, how many onto functions are there? We want all yi in the range of the function f. Call Pi the property that yi is not in the range of the the function f. Then we are looking for the number of functions that has none of the properties P1,...,Pn x y f A B There is no element without incoming arrows

Number of Onto-Functions N(P1’P2’P3’) = N-N(P1)-N(P2) ... + N(P1,P2) -...+(-1)^n N(P1,...,Pn). N: number of function from A  B: n^m N(Pi): number of functions that do not have y1 in its range: (n-1)^m. There are n=C(n,1) such terms. N(Pi,Pj): (n-2)^m with C(n,2) terms. Total: n^m – C(n,1)(n-1)^m + C(n,2)(n-2)^m – ...+(-1)^(n-1)C(n,n-1)1^m. m=6 and n=3 N(P1’P2’P3’) = 3^6 – C(3,1)2^6 + C(3,2)1^6 = 540 x y f A B There is no element without incoming arrows