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notes-090901 - MATH 681 1 Notes Combinatorics and Graph...

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MATH 681 Notes Combinatorics and Graph Theory I 1 The Inclusion-Exclusion Principle Our next step in developing the twelvefold way will deal with the surjective functions. We’ll build these through the use of inclusion-exclusion . In its most basic form, inclusion-exclusion is a way of counting the membership of a union of sets. For two sets, it is easy to cinvince yourself that | A B | = | A | + | B | - | A B | . With a little bit more doing, we can show that | A B C | = | A | + | B | - | A B | - | A C | - | B C | + | A B C | . We add and subtract certain slices of sets until all overcounts and undercounts are eliminated. We can use these simple, small versions of the inclusion-exclusion principle in a simple example: Question 1: How many members of { 1 , 2 , 3 , . . . , 105 } have nontrivial factors in common with 105? Answer 1: 105 = 3 · 5 · 7 , so a nubmer shares factors with 105 if and only if it is divisible by 3, 5, or 7. Let A , B , and C be the members of { 1 , 2 , 3 , . . . , 105 } divisible by 3, 5, and 7 respectively. Clearly | A | = 35 , | B | = 21 , and | C | = 15 . Furthermore, A B consists of those numbers divisible by both and 5, i.e., divisible by 15. Likewise, A C and B C contain multiples of 21 and 35 respectively, so | A B | = 7 , | A C | = 5 , and | B C | = 3 . Finally, A B C consists only of the nubmer 105 , so it has 1 member total. Thus, | A B C | = 35 + 21 + 15 - 7 - 5 - 3 + 1 = 57 There are 3 simple presentations of the inclusion-exclusion principle, which you should be able to convince yourself are equivalent: | A 1 A 2 ∪ · · · ∪ A n | = | A 1 | + | A 2 | + · · · + | A n | - | A 1 A 2 | - | A 1 A 3 | - · · · - | A 1 A n | - | A 2 A 3 | - · · · - | A n - 1 A n | + | A 1 A 2 A 3 | + | A 1 A 2 A 4 | + · · · + | A n - 2 A n - 1 A n | - · · · ± | A 1 A 2 ∩ · · · ∩ A n | All those ellipses are a bit unhappy, so one of the following forms is a bit more explicit if a bit less readable: | A 1 A 2 ∪ · · · ∪ A n | = n X i =1 ( - 1) i - 1 X | S | = i ; S ⊆{ 1 , 2 ,...,n } \ j S A j which can be simplified a bit further, by choosing a set first and then worrying about its size: | A 1 A 2 ∪ · · · ∪ A n | = X S ⊆{ 1 , 2 ,...,n } ; S 6 = ( - 1) | S |- 1 \ j S A j This form is also the easiest in which to prove the inclusion-exclusion principle. Page 1 of 7 September 1, 2009

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MATH 681 Notes Combinatorics and Graph Theory I 1.1 Proof of Inclusion-Exclusion Proposition 1. For finite sets A 1 , A 2 , . . . , A n , | A 1 A 2 ∪ · · · ∪ A n | = X S ⊆{ 1 , 2 ,...,n } ; S 6 = ( - 1) | S |- 1 \ j S A j Proof. We prove this by induction on n . For n = 1, it is trivial: | A 1 | = X ∅6 = S ⊆{ 1 } ( - 1) | S |- 1 \ j S A j For our inductive step, we will take it as given that: | A 1 A 2 ∪ · · · ∪ A n - 1 | = X S ⊆{ 1 , 2 ,...,n - 1 } ; S 6 = ( - 1) | S |- 1 \ j S A j and thereby find | A 1 A 2 ∪ · · · ∪ A n | : | A 1 ∪ · · · ∪ A n | = | ( A 1 A 2 ∪ · · · ∪ A n - 1 ) A n | = | A 1 A 2 ∪ · · · ∪ A n - 1 | + | A n | - | ( A 1 A 2 ∪ · · · ∪ A n - 1 ) A n | = | A 1 A 2 ∪ · · · ∪ A n - 1 | + | A n | - | ( A 1 A n ) ( A 2 A n ) ∪ · · · ∪ ( A n - 1 A n ) | = X S ⊆{ 1 , 2 ,...,n - 1 } ; S 6 = ( - 1) | S |- 1 \ j S A j + | A n | - X S ⊆{ 1 , 2 ,...,n - 1 } ; S 6 = ( - 1) | S |- 1 \ j S A j A n = X S ⊆{ 1 , 2 ,...,n - 1 } ; S 6 = ( - 1) | S |- 1 \ j S A j + | A n | - X S ⊆{ 1 , 2 ,...,n
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