# sol5 - MATH 239 Assignment 5 - Suggested Solutions 1(a) [1...

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MATH 239 Assignment 5 - Suggested Solutions 1(a) [1 mark] The vertices of A n are given by the n -element subsets of { 1 ,..., 2 n + 1 } , so the number of vertices in A n is given by p = p 2 n + 1 n P , n 1 . (b) [1 mark] For any vertex α , the neighbours are given by the n -element subsets of the set of n +1 elements not in α , so α has degree ( n +1 n ) = n +1, n 1. Thus A n is an ( n +1)-regular graph, and from the Handshake Theorem, the number of edges in A n is given by q = 1 2 p ( n + 1) = n + 1 2 p 2 n + 1 n P , n 1 . (c) [2 marks] Suppose that a triangle exists in A n , with vertices α,β,γ . Then α,β,γ are all disjoint from each other (i.e., α β = α γ = β γ = ). Then S = α β γ is a subset of { 1 ,..., 2 n + 1 } with | S | = 3 n , which implies that 3 n 2 n + 1. Subtracting 2 n from each side of this inequality, we get n 1. We conclude that A n cannot contain a triangle if n > 1. When n = 1, A n does indeed contain the triangle { 1 } , { 2 } , { 3 } . 2(a)

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## sol5 - MATH 239 Assignment 5 - Suggested Solutions 1(a) [1...

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