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.
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 $m=n.$