Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The 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
$$\sum_{k=0}^r \binom{m(k)}{k}\binom{n(k)}{r-k}=\binom{m+n}{r}?$$

share|improve this question

1 Answer 1

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.

share|improve this answer
    
thanks a lot! These identities are nice! –  pipi Nov 28 '12 at 1:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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