Edge count of a complete graph ($|E(K_n)|=\left(\genfrac{}{}{0ex}{}{n}{2}\right)=n(n-1)/2$)

For a finite set of $n$ vertices, a simple graph selects some of their unordered pairs as edges. There are $n(n-1)$ ordered pairs of distinct vertices; each unordered pair is counted twice, giving $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ possibilities. Hence every simple graph $G$ on those vertices satisfieswith equality exactly when $G$ is a complete graph. The binomial coefficient also shows why the count grows quadratically with the number of vertices.