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Part 12 Inclusion-Exclusion and Applications Printed version of the lecture Discrete Mathematics on 7. October 2009 Tommy R. Jensen, Department of Mathematics, KNU 12.1 Contents 1 The Inclusion-Exclusion Principle 1 2 Applications of the Inclusion-Exclusion Principle 3 3 Derangements 4 4 Conclusion 5 12.2 1 The Inclusion-Exclusion Principle Inclusion-exclusion for more than two sets 12.3 1 The general inclusion-exclusion principle Notation Let X be any set and k an integer. Then we can write X = {Y ⊆ X : |Y | = k}, k for the set of k-subsets of X. Theorem 6.1.1 (inclusion-exclusion principle) Let A1 , A2 , . . . , Am be subsets of a finite set S. Then |A1 ∩ A2 ∩ · · · ∩ Am | = |S| + m (−1)k ∑ |Ai1 ∩ Ai2 ∩ · · · ∩ Aik |. {i1 ,i2 ,...,ik }∈({1,2,...,m} ) k ∑ k=1 Example. If A, B,C are subsets of S, then |A ∩ B ∩C| = |S| − |A| − |B| − |C| + |A ∩ B| + |A ∩C| + |B ∩C| − |A ∩ B ∩C|. 12.4 Inclusion-exclusion principle in action Example. |A ∩ B ∩C| = |S| − |A| − |B| − |C| + |A ∩ B| + |A ∩C| + |B ∩C| − |A ∩ B ∩C|. 12.5 Proof of m |A1 ∩ A2 ∩ · · · ∩ Am | = |S| + (−1)k ∑ |Ai1 ∩ Ai2 ∩ · · · ∩ Aik |. {i1 ,i2 ,...,ik }∈({1,2,...,m} ) k ∑ k=1 How much does an element of S contribute on the left side? It contributes 1 if it is not an element of any set Ai . Otherwise it contributes 0. 12.6 2 m (−1)k ∑ |Ai1 ∩ Ai2 ∩ · · · ∩ Aik |. {i1 ,i2 ,...,ik }∈({1,2,...,m} ) k How much does an element x of S contribute on the right side? First we assume that x is in some set Ai , and prove that x contributes 0 to the right side. |A1 ∩ A2 ∩ · · · ∩ Am | = |S| + ∑ k=1 Let A j1 , A j2 , . . . , A jn be the subsets that contain x, so 1 ≤ n ≤ m. There is a contribution of 1 from x to |Ai1 ∩ Ai2 ∩ · · · ∩ Aik | if and only if {i1 , i2 , . . . , ik } ⊆ { j1 , j2 , . . . , jn }. Because of the factor (−1)k the contribution is +1 if k is even and −1 if k is odd. The number of even and odd subsets of { j1 , j2 , . . . , jn } are equal, and the contribution from x is −1 from every odd subset and +1 from every even subset except 0/ (because k > 0). We deduce that x contributes +1 to |S| and −1 to the summation, so altogether x contributes 0 on the right side. 12.7 m |A1 ∩ A2 ∩ · · · ∩ Am | = |S| + (−1)k ∑ |Ai1 ∩ Ai2 ∩ · · · ∩ Aik |. {i1 ,i2 ,...,ik }∈({1,2,...,m} ) k ∑ k=1 How much does an element x of S contribute on the right side? Now we assume that x is not in any set Ai , and prove that x contributes 1 to the right side. But this is clear, because x contributes +1 to |S| and 0 to the summation of the sizes of intersections of the sets A1 , A2 , . . . , Am . This proves the Theorem. 12.8 Another form of the principle Corollary 6.1.2 Let A1 , A2 , . . . , Am be subsets of a finite set S. Then m |A1 ∪ A2 ∪ · · · ∪ Am | = ∑ k=1 (−1)k+1 ∑ |Ai1 ∩ Ai2 ∩ · · · ∩ Aik |. {i1 ,i2 ,...,ik }∈({1,2,...,m} ) k Proof of Corollary 6.1.2 |A1 ∪ A2 ∪ · · · ∪ Am | = |S| − |A1 ∩ A2 ∩ · · · ∩ Am |. Now use Theorem 6.1.1. 2 12.9 Applications of the Inclusion-Exclusion Principle Example How many of the integers 1, 2, 3, . . . , 2009 are not divisible by any of the numbers 4, 5, 6? Answer: Let S = {1, 2, 3, . . . , 2009}. A1 = {4, 8, 12, . . . , 2008}, |A1 | = b 2009 4 c = 502. A2 = {5, 10, 15, . . . , 2005}, |A2 | = b 2009 5 c = 401. A3 = {6, 12, 18, . . . , 2004}, |A3 | = b 2009 6 c = 334. The least common multiple of some numbers a1 , a2 , . . . , an is the smallest positive number which is divisible by all of a1 , a2 , . . . , an . It is written as lcm{a1 , a2 , . . . , an }. A1 ∩A2 contains exactly numbers divisible by 4 and 5. These are precisely the numbers {20, 40, 60, . . . , 2000} divisible by lcm{4, 5} = 20. Therefore |A1 ∩ A2 | = b 2009 20 c = 100. We get lcm{4, 6} = 12, so |A1 ∩ A3 | = b 2009 c = 167. 12 And lcm{5, 6} = 30 implies |A2 ∩ A3 | = b 2009 30 c = 66. 3 Finally lcm{4, 5, 6} = 60, and |A1 ∩ A2 ∩ A3 | = b 2009 60 c = 33. From Theorem 6.1.1: |A1 ∩ A2 ∩ A3 | = 2009 − 502 − 401 − 334 + 100 + 167 + 66 − 33 = 1072. 12.10 3 Derangements Derangements Definition A derangement of the set {1, 2, . . . , n} is a permutation i1 i2 . . . in of {1, 2, . . . , n} such that no number is in its own place: i1 6= 1, i2 6= 2, . . . , in 6= n. From the online Wiktionary: Adjective deranged (comparative more deranged, superlative most deranged) • 1. disturbed or upset, especially mentally • 2. insane Synonyms * See also Wikisaurus:insane 12.11 The number of derangements Theorem 6.3.1 The number of derangements of {1, 2, . . . , n} is equal to 1 1 1 1 Dn = n! 1 − + − + · · · + (−1)n . 1! 2! 3! n! 12.12 Proof of Theorem 6.3.1 Let S be the set of all permutations of {1, 2, . . . , n} : |S| = n! For 1 ≤ j ≤ n let A j be the permutations i1 i2 . . . in with i j = j. Then Dn = |A1 ∩ A2 ∩ · · · ∩ An |. What are the permutations in A1 ∩ A2 ∩ · · · ∩ Ak ? They are of the form i1 i2 . . . ik ik+1 . . . in = 12 . . . kik+1 . . . in , where ik+1 . . . in is any permutation of {k + 1, . . . , n}. Since there are (n − k)! permutations of {k + 1, . . . , n} we get |A1 ∩ A2 ∩ · · · ∩ Ak | = (n − k)! We obtain the same number (n − k)! of permutations that place any k numbers i1 , i2 , . . . , ik each in its own place. 12.13 Proof of Theorem 6.3.1 Theorem 6.1.1 can now be used, with |S| = n! and |Ai1 ∩ Ai2 ∩ · · · ∩ Aik | = (n − k)! for all i1 , i2 , . . . , ik : n (−1)k ∑ |Ai1 ∩ Ai2 ∩ · · · ∩ Aik | k=1 {i1 ,i2 ,...,ik }∈({1,2,...,n}) k n n k n k (n − k)! = n! + ∑ (−1) ∑(n − k)! = n! + ∑ (−1) k k=1 {i1 ,i2 ,...,ik }∈({1,2,...,n}) k=1 k ! n n k 1 k n! = n! 1 + ∑ (−1) = n! + ∑ (−1) k! k! k=1 k=1 1 1 1 1 = n! 1 − + − + · · · + (−1)n . 1! 2! 3! n! Dn = |S| + ∑ 12.14 4 4 Conclusion Conclusion This ends the lecture! 12.15 Next time: Generating Functions 12.16 5