Sum over k of r-k Choose m by s Choose k-t by -1^k-t

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Theorem

Let $s \in \R, r, t, m \in \Z_{\ge 0}$.

Then:

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

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


Proof

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


Sources