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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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    Perhaps, you can add a new question with the original problem and link that to this one. You might get an answer which does not use this hint at all...2010-12-28
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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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    Yes, I tried that but the first expression is not really a $S(r,n)$ because the coefficient $\frac{m}{m+v}$ changes with the sum.2010-12-27
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    @Rob: I said the case $m=0$, which leaves us with the second expression. Even in special case of $m=0$, you get $S(r,n)$, so with $m \neq 0$, it will certainly be more difficult!2010-12-27
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    @Rob: In case you were responding to the last sentence of the answer...I would suggest looking that the proofs of the identity with $S(n,r)$ I gave. Perhaps you can modify one of those to cater to your $m/(m+v)$.2010-12-27
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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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    It looks like you can use $\frac{m}{m+v}-1=\frac{-v}{m+v}$ to combine the terms2010-12-27
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    @Ross: That's a good idea, but what can I do with term with $\frac{v}{m+v}$? Supposedly, I should be able to find a single expression.2010-12-27
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    That gets you to a single sum. You can take out the v against the $\binom{n-m}{v}$ but that doesn't seem to help. If I could go farther, I would have posted an answer instead of a comment.2010-12-27
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    @Ross: Right. It could have been helpful, but not for this particular problem.2010-12-27