Binomial sum with two parameters Let $m$ and $n$ be two integers. Evaluate 
$$S_{m,n}=\sum_{j=0}^{m} (-1)^j \binom{m}{j}\binom{mn-jn}{m+1}$$
At first, for $n=2$ I got $S_{m,2}=2^{m-1}m$, for $n=3$ I obtained $S_{m,3}=3^m m$, then I tried in vain to prove by induction that
$$S_{m,n}=\frac{n^k k(n-1)}{2}$$ 

Suppose we  seek to evaluate
$$S(m,n) = \sum_{j=0}^m (-1)^j {m\choose j}
{mn-jn\choose m+1}.$$
Introduce
$${mn-jn\choose m+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn-jn} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn}
\sum_{j=0}^m 
(-1)^j {m\choose j} \frac{1}{(1+z)^{jn}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn}
\left(1-\frac{1}{(1+z)^n}\right)^m \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} ((1+z)^n-1)^m \; dz.$$
This is
$$[z^{m+1}] ((1+z)^n-1)^m
= [z^{m+1}] \left(nz+{n\choose 2}z^2+\cdots\right)^m.$$
This evaluates by inspection to
$${m\choose 1} n^{m-1} {n\choose 2}
= \frac{1}{2} n^m (n-1) m$$
because from the  $m$ terms being multiplied we must  choose one to be
$z^2$ to  ge a total of  $m+1$ and there  are $m-1$ terms in  $z$ with
coefficient $n.$
Observe that this formula produces zero when $n=1$ which is correct as well (second binomial coefficient is zero).
