TITLE: a sum with binomial coefficients
QUESTION [19 upvotes]: Let integers $n,k$ satisfy $0 \le k \le n$.  We desire proof that
$$
{n\choose k} =
\sum {n\choose a}(-1)^a\;{-k\choose b}(-1)^b\;{-(n-k)\choose c}(-1)^c
\tag{$*$}$$
where the (finite) sum is over all ordered triples $(a,b,c)$ of nonnegative integers satisfying
$$
a\cdot n + b\cdot k + c\cdot(n-k) = k(n-k) 
\\
\text{(or equivalently}\quad
\frac{a+b}{n-k} + \frac{a+c}{k} = 1 \quad).
$$
Of course the binomial coefficient for a binomial to a negative power is
$$
{-k\choose b} = \frac{(-k)(-k-1)(-k-2)\cdots(-k-b+1)}{b!} .
$$
Now two of the terms in ($*$)
are $(a,b,c)=(0,n-k,0)$ and $(a,b,c)=(0,0,k)$.  Those two terms do, indeed, add to ${n\choose k}$.  But then the problem becomes: show that the rest of the terms sum to zero.
Note
This arose from my attempt to solve 
integral of a "sin-omial" coefficients=binomial .
It is the calculation of the residue at $w=0$ of rational function
$$
\frac{(w^n-w^{-n})^n}{(w^k-w^{-k})^k(w^{n-k}-w^{-(n-k)})^{n-k}w}
$$
obtained by changing variables in the integral of that problem.

REPLY [2 votes]: This is an elementary/self-contained proof based on manipulation of
  binomials.
Let $l=n-k$ and $g=\gcd(n,k)=\gcd(k,l)$.
The condition $an+bk+cl=kl$ (where $a$, $b$, and $c$ are as in the
  original problem description) is equivalent to $(a+c)n=k(l+c-b)$,
  which means $k/g\mid a+c$, so $a+c=tk/g$, $a+b=l(1-t/g)$, and the
  only possible values of $t$ are $0,1,\ldots,g$. $t=0$ and $t=g$
  correspond to $(a,b,c)=(0,l,0)$ and $(a,b,c)=(0,0,k)$ respectively,
  which provide ${n\choose k}$ as mentioned
  in the problem description (by an easy binomial sum).
The problem is to show that $t=1,\ldots,g-1$ contribute zero in
  total to the sum (obvious if $n$ and $k$ are coprime). Helpfully it
  turns out that the contributions from each value of $t$ are
  separately zero, not just in aggregate, and the WZ method can be
  applied to each $t$ as follows.
To make the notation slightly prettier, define $\tau=t/g$ and
  $\tau'=1-t/g$. (There is then a symmetry $k\leftrightarrow l$,
  $\tau\leftrightarrow\tau'$.) $\tau$ and $\tau'$ will only ever
  occur multiplied by $k$, $l$ or $n$, so everything is an integer.
What needs to be proved is, for fixed $t$ between $1$ and $g-1$,
  $$\sum_{a=0}^{\min(k\tau,l\tau')} (-1)^a {n\choose a}{-k\choose l\tau'-a}{-l\choose k\tau-a}=0.$$
This is established by the explicit sum, valid for $0<\tau<1$:
  $$\sum_{a=0}^{r-1} (-1)^a {n\choose a}{-k\choose l\tau'-a}{-l\choose k\tau-a}=
  (-1)^{r+1} {n\choose r}{-k\choose l\tau'-r}{-l\choose k\tau-r}
  \frac{r(l+k\tau-r)(k+l\tau'-r)}{kln\tau\tau'},$$
  which vanishes for $r>\min(k\tau,l\tau')$. It is routine to check
  that the delta of the RHS is the LHS.