Each vertex in \(K_n\) is adjacent to \(n-1\) other vertices. We call the number of edges emanating from a given vertex the degree of that vertex. So every vertex in \(K_n\) has degree \(n-1\text{.}\) How many edges does \(K_n\) have? One might think the answer should be \(n(n-1)\text{,}\) since we count \(n-1\) edges \(n\) times (once for each vertex). However, each edge is incident to 2 vertices, so we counted every edge exactly twice. Thus there are \(n(n-1)/2\) edges in \(K_n\text{.}\) Alternatively, we can say there are \({n \choose 2}\) edges, since to draw an edge we must choose 2 of the \(n\) vertices.
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