Sum over k of r-k Choose m by s+k Choose n

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Theorem

Let $m, n, r, s \in \Z_{\ge 0}$ such that $n \ge s$.

Then:

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

where $\dbinom {r - k} m$ etc. are binomial coefficients.


Proof

\(\displaystyle \) \(\) \(\displaystyle \sum_{k \mathop = 0}^r \binom {r - k} m \binom {s + k} n\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop = 0}^r \binom {-\left({m + 1}\right)} {r - k - m} \binom {-\left({n + 1}\right)} {s + k - m} \left({-1}\right)^{r - m + s - m}\) Moving Top Index to Bottom in Binomial Coefficient
\(\displaystyle \) \(=\) \(\displaystyle \binom {-\left({m + 1}\right) - \left({n + 1}\right)} {r - m + s - m} \left({-1}\right)^{r - m + s - m}\) Chu-Vandermonde Identity
\(\displaystyle \) \(=\) \(\displaystyle \binom {r + s + 1} {m + n + 1}\) Moving Top Index to Bottom in Binomial Coefficient

$\blacksquare$


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