The proof proceeds by induction over $m$. For all $m \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^0 \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k}       | r = \dbinom r 0 \dbinom s {n - 0} \paren {n r - \paren {r + s} 0}       | c =  }} {{eqn | r = \dbinom s n \paren {n r}       | c = Binomial Coefficient with Zero }} {{eqn | r = \paren {0 + 1} \paren {n - 0} \dbinom r {0 + 1} \dbinom s {n - 0}        | c = Binomial Coefficient with One }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({j}\right)$ is true, where $j \ge 1$, then it logically follows that $P \left({j + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{k \mathop = 0}^j \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {j + 1} \paren {n - j} \dbinom r {j + 1} \dbinom s {n - j}$ from which it is to be shown that: :$\displaystyle \sum_{k \mathop = 0}^{j + 1} \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {j + 2} \paren {n - j - 1} \dbinom r {j + 2} \dbinom s {n - j - 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | r = \sum_{k \mathop = 0}^{j + 1} \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k}       | o =        | c =  }} {{eqn | r = \sum_{k \mathop = 0}^j \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} + \dbinom r {j + 1} \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} }       | c =  }} {{eqn | r = \paren {j + 1} \paren {n - j} \dbinom r {j + 1} \dbinom s {n - j} + \dbinom r {j + 1} \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} }       | c =  }} {{eqn | r = \dbinom r {j + 1} \paren {\paren {j + 1} \paren {n - j} \dbinom s {n - j} + \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} } }       | c = factorising }} {{eqn | r = \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \paren {\paren {j + 1} \paren {n - j} \dbinom s {n - j} + \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} } }       | c = Factors of Binomial Coefficient: Corollary 2 }} {{eqn | r = \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \paren {\dfrac {\paren {j + 1} \paren {n - j} } {\paren {n - j} \paren {s - \paren {n - j} } } \dbinom s {n - j - 1} + \dbinom s {n - j - 1} \paren {n r - \paren {r + s} \paren {j + 1} } }       | c = Factors of Binomial Coefficient: Corollary 2 }} {{eqn | r = \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \dbinom s {n - j - 1} \paren {\dfrac {j + 1} {s - n + j} + \paren {n r - \paren {r + s} \paren {j + 1} } }       | c = simplifying }} {{eqn | r = \paren {j + 2} \paren {r - j + 2} \dbinom r {j + 2} \dbinom s {n - j - 1} \paren {\dfrac {j + 1} {s - n + j} + \paren {n r - r j - s j - r - s} }       | c = simplifying }} {{end-eqn}} {{finish|It's not working out as neatly as I'd hoped. Is there something simple I'm missing?}} So $P \left({j}\right) \implies P \left({j + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$