Summation from k to m of r Choose k by s Choose n-k by nr-(r+s)k

Theorem

 * $\displaystyle \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$

Proof
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:

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:

So $P \left({j}\right) \implies P \left({j + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \sum_{k \mathop = 0}^m \dbinom r k \dbinom s {n - k} \paren {n r - \paren {r + s} k} = \paren {m + 1} \paren {n - m} \dbinom r {m + 1} \dbinom s {n - m}$