# What is the average degree of a complete graph and why?

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.

• Each vertex has exactly $n - 1$ neighbors, thus $n - 1$. Mar 7 '16 at 3:20

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 $n-1$.