= Solution
Count pairs $(M,K)$ of <subsets> of an $n$-element candidate <set> with $K\subseteq M$, $|M|=m$, and $|K|=k$. Choosing the whole team $M$ first and then the special-role players $K$ gives $\binom nm\binom mk$. Alternatively choose $K$ first, then choose the other $m-k$ players from the $n-k$ candidates outside $K$, giving $\binom nk\binom{n-k}{m-k}$. Each construction specifies the same pair uniquely. Thus <double counting> proves
$$
\boxed{\binom nm\binom mk=\binom nk\binom{n-k}{m-k}.}
$$
Back to article page
