# Looking for a proof of an interesting combinatorial identity

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 $$\nu,\mu$$?

• 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. – darij grinberg Apr 14 at 23:08
• @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? – user Apr 14 at 23:13

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).