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

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Unformatted text preview: 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 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 ∈...
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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.

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

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