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Unformatted text preview: MATH 681 Notes Combinatorics and Graph Theory I 1 The InclusionExclusion Principle Our next step in developing the twelvefold way will deal with the surjective functions. We’ll build these through the use of inclusionexclusion . In its most basic form, inclusionexclusion 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 inclusionexclusion 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 inclusionexclusion 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 inclusionexclusion principle. Page 1 of 7 September 1, 2009 MATH 681 Notes Combinatorics and Graph Theory I 1.1 Proof of InclusionExclusion 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 ∈...
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This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.
 Fall '09
 WILDSTROM
 Combinatorics, Graph Theory

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