Let $j, k ,n$ be nonnegative integers such that $0 \leq j, k \leq n \leq k +j $. Pick integer $m$ such that $0 \leq m \leq k + j - n$.


Let $\langle x \rangle\_m$ denote the falling factorial $x(x-1)\ldots (x-m+1)$.


I've stumbled across the need to prove the following equality:


$$\sum\_{i=0}^n \frac{\binom{k}{i}\binom{j}{i}}{\binom{n}{i}}(-1)^i\langle ix\rangle\_m = \left\{ \begin{matrix} \frac{k! j!}{n!} (-x)^m ~~~~~~~~\text{when }m = k+j - n\\ 0 ~~~~~~~~~~~~~~~~~\text{when }m < k + j - n \end{matrix} \right.$$
An equivalent formulation is:
$$\sum\_{i=0}^n \frac{(-1)^i\binom{k}{i}\binom{j}{i} \binom{i x}{m}}{\binom{n}{i}}= \left\{ \begin{matrix} \frac{k! j!}{n!m!} (-x)^m ~~~~~~~~\text{when }m = k+j - n\\ 0 ~~~~~~~~~~~~~~~~~\text{when }m < k + j - n \end{matrix} \right.$$
Any references to potentially related material would be greatly appreciated.


${\scriptsize \textbf{Edited to fix sign}}$



Combinatorial Identity with Connection Coefficients and Falling Factorial $\langle i x\rangle_n$
It is very probable that what is written below is the simplification of Darij's argument. I use the notation $x^{\underline{n}}=x(x-1)\dots(x-n+1)$ [as in Knuth's books] for the falling factorial, and $[t^n] f(t)$ for the coefficient of $t^n$ in the polynomial $f$.


For a polynomial $f(t)$ of degree at most $j$ we have $$\sum\_{i=0}^j (-1)^{j-i}\frac{f(i)}{i!(j-i)!}=[t^j]f(t),$$
this follows from the Lagrange interpolation of $f$ in the points $\{0,1,\dots,j\}$. Apply this to the polynomial $f(t)=(tx)^{\underline m}(n-t)^{\underline {n-k}}$. In RHS we have 0 if $m<k+j-n$ and $(-1)^{n-k}x^m$ if $m=k+j-n$. In LHS we get $$\sum\_{i=0}^{\min(j,k)} 
(-1)^{j-i}\frac{(n-i)!(ti)^{\underline m}}{(k-i)!i!(j-i)!}=(-1)^j\frac{n!}{k!j!}\sum\_{i=0}^{\min(j,k)} 
(-1)^{i}\frac{\binom{k}i\binom{j}i}{\binom{n}i}(ti)^{\underline m}$$
and we are done.


