Prove that if a graph of order 3n (n ≥ 1) has n vertices of each of the degrees n − 1, n and n + 1, then n is even.
A problem from Graph Theory.
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graph-theory
1 Answers
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For any graph , the number of edges is half the sum of degrees of all vertices, and must be a whole number. On calculation we find $ E = \frac{3n^2}{2} $ . Thus 2 divides n and the number of vertices is even.