Applying the basic definition gives

$$\begin{array}{rll} \displaystyle\binom{n}{k}\binom{k}{m}&=\displaystyle\frac{n!}{k!(n-k)!}\cdot\frac{k!}{m!(k-m)!} &=\displaystyle\frac{n!}{(n-k)!m!(k-m)!} \\ \displaystyle\binom{n}{m}\binom{n-m}{k-m} &= \displaystyle\frac{n!}{m!(n-m)!}\cdot\frac{(n-m)!}{(k-m)!(n-k)!} &=\displaystyle\frac{n!}{(n-k)!m!(k-m)!} \\ \displaystyle\binom{n}{k-m}\binom{n-k+m}{m}&=\displaystyle \frac{n!}{(k-m)!(n-k+m)!}\cdot\frac{(n-k+m)!}{m!(n-k)!} &=\displaystyle\frac{n!}{(n-k)!m!(k-m)!}\end{array}$$

They are all equal.