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Sorry for this uninteresting question but hopefully someone can provide some help.

Is there a way to simplify the following expression?

$\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}\displaystyle \frac{m}{m+v}$

$-\binom{n}{m}\sum_{\nu=0}^{n-m}(-1)^{\nu}\binom{n-m}{v}\left(\frac{n-m-v}{n}\right)^{r}$

This is a hint for a problem, but I don't know how to proceed.

UPDATE: Would it bother you if I add the original problem?. Maybe some context is needed in order to solve this problem by using this hint.

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    @Moron: Ok, I'll do that.2010-12-28

2 Answers 2

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I doubt it, the case $m=0$ would give a simpler formula for Stirling numbers of the second kind.

$n!\ S(r,n) = \sum_{v = 0}^{n} (-1)^{v} {n \choose v} (n-v)^{r}$

Perhaps you can write your expression in terms of these numbers...

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    Uhm, yes, I was looking for an identity but didn't find any. I suppose this shouldn't be relying in some obscure identity (if there is any which fits the requirements).2010-12-27
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At a quick glance, it looks like there are some common factors to the two sums and that factoring is likely to help in simplifying. Also, since the summations have the same indexing and range, they could be combined into a single summation.

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    @Ross: Right. It could have been helpful, but not for this particular problem.2010-12-27