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Trying to generalize a combinatorial identity I have ended up with the following expression: $$ \sum_{\{k\}_L}\binom\nu K K!\prod_l \frac1{k_l!}\binom \mu l^{k_l}=\binom{\nu\mu}L,\tag1 $$ where the sum runs over all sets $\{k\}_L$ of integer numbers $k_l$ $(l\ge1)$ satisfying $$k_l\ge0,\;\sum_{l\ge1} lk_l=L,\tag2$$ and $K$ is alias for $\sum_l k_l$.

The expression was derived by a combinatorial argument for positive integer $\nu,\mu$. However by numerical evidence it seems to be valid for arbitrary complex numbers as well (with $\Gamma$-function used in place of factorials).

Is the expression $(1)$ known? How can it be algebraically proved for general complex $\nu,\mu$?

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  • $\begingroup$ Which numbers are you making complex? $\nu$? $\mu$? You don't need the Gamma-function for that; both sides are polynomials in $\nu$ and $\mu$, so the general case follows automatically once you know that the identity holds for nonnegative integers $\nu$ and $\mu$. On the other hand, I have no idea how to make $L$ complex; I suspect that's not what you want. $\endgroup$ – darij grinberg Apr 14 at 23:08
  • $\begingroup$ @darijgrinberg Yes it concerns only $\nu $ and $\mu $. I thought I have expressed it clear enough, but if it is not the case, I am very sorry. Could you please turn your comment into an answer? $\endgroup$ – user Apr 14 at 23:13
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As you said in the comment, you are only trying to replace $\nu$ and $\mu$ (but not $L$) by complex numbers. You do not need the $\Gamma$-function for that; a binomial coefficient of the form $\dbinom{x}{k}$ is defined in the usual way (namely, as $\dfrac{x\left(x-1\right)\cdots\left(x-k+1\right)}{k!}$) whenever $x$ is a complex number and $k$ is a nonnegative integer. (The $\Gamma$-function only becomes necessary if you want to extend it nontrivially to non-integer values of $k$; even then, it is not clear whether such an extension is the most useful one.)

Furthermore, your identity (1) is an equality between two polynomials in $\nu$ and $\mu$ (when $L$ is held constant). The "polynomial identity trick" says that if such an equality holds whenever $\nu$ and $\mu$ are nonnegative integers, then it must also hold for arbitrary complex numbers $\nu$ and $\mu$. Since you (presumably) have a proof of (1) in the case whenever $\nu$ and $\mu$ are nonnegative integers, you thus automatically obtain a proof of (1) for arbitrary complex numbers $\nu$ and $\mu$. No further argument is required (though, of course, you can choose to come up with a different proof).

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