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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 $n - 1$. $\endgroup$ – vonbrand Mar 7 '16 at 3:20
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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 $n-1$.

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