A combinatorial identity generalizing identity (3.111) from Gould's book I am trying to prove the following identity, which I am sure is correct:
$$
\sum_{m=0}^{K}(-1)^m{n \choose m}{(n-m)r-1 \choose n-1}=1,
$$
where $K:=\left[n\frac{r-1}{r}\right]$ for some integer $r$, and $[x]$ denotes the largest integer contained in $x$.
In the special case $r=2$ this is identity (3.111) from H. Gould's book "Combinatorial identities" (1972). For $r=2$ it was first stated by B. C. Wong in Amer. Math. Monthly in May 1930 as an open question. I am wondering if someone has seen a proof of (or can prove) this identity for any $r$. The upper summation limit$\left[n\frac{r-1}{r}\right]$ also occurs in Gould's book in identity (3.113), so it is not entirely unheard of.
 A: This  one can  also be  done using  complex variables.

Suppose we are trying to evaluate
$$\sum_{m=0}^{\lfloor n(r-1)/r\rfloor} {n\choose m} (-1)^m
{(n-m)r-1\choose n-1}.$$
Introduce the integral representation
$${(n-m)r-1\choose n-1}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{(n-m)r-1}}{z^{n}} \;dz.$$
This integral  sets the range  of the sum, ensuring  that $(n-m)r-1\ge
n-1$ or $(n-m)r\ge  n$ or $n(r-1)\ge mr$ or  $n(r-1)/r\ge m,$ which is
precisely the definition of the constant $K,$ except when $n=m$, where
it yields 
$$\frac{1}{(n-1)!}
(-1)\times(-2)\times(-3)\times\cdots\times (-(n-1))
= (-1)^{n-1}$$
so we can let $m$ range up to $n$ to obtain
$$-(-1)^{n-1}\times (-1)^n + \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{nr-1}}{z^n}
\sum_{m=0}^n {n\choose m} (-1)^m
\frac{1}{(1+z)^{mr}} \; dz.$$
This is
$$(-1)^{2n} + \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{nr-1}}{z^n}
\left(1-\frac{1}{(1+z)^r}\right)^n \; dz$$
or
$$1 + \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^n(1+z)}
\left((1+z)^r-1\right)^n \; dz$$
This becomes
$$1+[z^{n-1}] \frac{((1+z)^r-1)^n}{1+z}.$$
Note however that $((1+z)^r-1)^n$ starts at $z^n$ and $1/(1+z)$ starts
at $z^0$, so the contribution from the integral is zero, leaving just
$$1$$
as the end result.

Remark.  When  $$n(r-1)/r <  m$$  with  $n\ne  m$ we  obtain  that
$(n-m)r-1 < n-1$ but $(n-m)r-1 \ge r-1$ is positive and hence $(1+z)^{(n-m)r-1}$ is a polynomial, producing the value zero for the binomial coefficient except when $m=n.$
