Is it correct that the average degree of a complete graph $K_n$, is $(n-1)/n$? or is it only $(n-1)$?
I have the answer of my professor which is $(n-1)$, but I didn't understand why.
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$\begingroup$ Each vertex has exactly $n - 1$ neighbors, thus $n - 1$. $\endgroup$– vonbrandMar 7, 2016 at 3:20
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1 Answer
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In a complete (undirected) graph with $n$ vertices, every vertex is connected to all other $n-1$ vertices, so the degree of every vertex is $n-1$. In particular, the average degree is $n-1$.
