Vandermonde's identity gives $$\sum_{k=0}^r \binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}.$$

Here is an example of Vandermonde's-like identity: For all $0 \le m \le n$, $$\sum_{k=0}^{2m} \binom{\left\lfloor\frac{n+k}{2}\right\rfloor}{k}\binom{m+\left\lfloor\frac{n-k}{2}\right\rfloor}{2m-k}=\binom{m+n}{2m}$$ (Note that $\left\lfloor\frac{n+k}{2}\right\rfloor+\left(m+\left\lfloor\frac{n-k}{2}\right\rfloor\right)$ is either $m+n$ or $m+n \pm 1$)

I wonder if there are some similar identities where $m(k)$ and $n(k)$ are functions of $k$ and $m(k)+n(k)$ is 'almost' constant, says $m+n$, the identity looks like
$$\sum_{k=0}^r \binom{m(k)}{k}\binom{n(k)}{r-k}=\binom{m+n}{r}?$$


Here are some almost "Vandermonde-like" identities that may be of interest. They're not exactly what you're asking for, but they are pretty close.

$$\begin{align*} \sum_{k=0}^n \binom{p+k}{k} \binom{q+n-k}{n-k} &= \binom{n+p+q+1}{n} \\ 2 \sum_{k=0}^r \binom{n}{2k} \binom{n}{2r+1-2k} &= \binom{2n}{2r+1} \\ 2 \sum_{k=0}^r \binom{n}{2k} \binom{n}{2r-2k} &= \binom{2n}{2r} + (-1)^k \binom{n}{r} \\ 2 \sum_{k=0}^{r-1} \binom{n}{2k+1} \binom{n}{2r-2k-1} &= \binom{2n}{2r} - (-1)^k \binom{n}{r} \end{align*}$$

The first one is on p. 148 of Riordan's Combinatorial Identities, and the last three are on p. 144. There may be more in Riordan's book; I just flipped through until I found a few.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.