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Let $p>d>0$ be given integers. Is there any trick how I can analytically calculate / simplify the following term:

$$\sum _{k=0}^d \frac{e^{-\frac{k^2}{2}} \binom{d}{k} \binom{p-d}{d-k}}{\binom{p}{d}}$$

Edit: Since this seems impossible I'm looking for a lower bound

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    $\begingroup$ Somehow seems unlike. The $e^{-k^2}$ is really problematic. $\endgroup$ May 14 at 0:53

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