# Cycles

I'm not a graph theorist, but I think since we are considering $K_{m,n}$ for $m,n \geq 2$, then any four distinct vertices -- 2 coming from the set of m vertices and 2 coming from the set of n vertices -- will make a 4-cycle. So that there will be $\left( \begin{array}{l} m \\ 2 \\ \end{array} \right) \left( \begin{array}{l} n \\ 2 \\ \end{array} \right)$ ways of choosing such a collection of 4 vertices.