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I recently came across the identity

$$\sum_{k=0}^m\dbinom{m}{k}\cdot \frac{(-1)^k}{n+k+1}=\dfrac{n!\cdot m!}{(n+m+1)!},$$

while working on evaluating

$$\int_0^1 x^n(1-x)^m\, dx.$$

I ended up showing that both sides of the identity were equal to this integral, but I was wondering if there was a way to show directly (either by manipulation or some combinatorial argument) that one was equal to the other. Any help is much appreciated.

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  • $\begingroup$ This identity appeared at this MSE link. $\endgroup$ – Marko Riedel Sep 4 '15 at 20:35
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One possible combinatorial explanation results from multiplying everything by $n+1$: $$ \sum_{k=0}^m \binom{m}{k} (-1)^k \frac{n+1}{n+1+k} = \frac{1}{\binom{n+m+1}{m}}. $$ On the left-hand side we have an inclusion-exclusion formula, which counts the probability that none of the bad events $B_1,\ldots,B_m$ happen; the probability that $k$ specific bad events happen is $(n+1)/(n+1+k)$.

One possible experiment which can underlie this identity is the one in which you sample $m$ balls with replacement out of $n+2$ "colors", with uniform distribution over the $\binom{n+m+1}{m}$ different choices. The bad event $B_i$ is that ball $i$ doesn't get color $1$, say. The probability that none of the bad events happen is exactly $1/\binom{n+m+1}{m}$. Each single balls is chosen uniformly, so the probability of $B_i$ is indeed $(n+1)/(n+2)$. I conjecture that the probability of the conjunction of any $k$ bad events is $(n+1)/(n+1+k)$, which implies your identity.

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