Prove of Vandermonde's identity

Question

I have been searched for the way of proving the following vandermonde’s identity for a long time which is far different from the original one, but I failed to find. I tried all of binomial theorem I know, and also used the way of combinatorial proof by setting several groups that has l-k indiv where m people are chosen and q+k indiv where n peole are chosen, but I have no idea where that “1” came from

$$ \sum_{k=0}^{l}{l-k\choose m}{q+k\choose n}= {l+q+1\choose m+n+1} \\(l,m,q,n >= 0)$$

How can i prove this identity? I wanna know both the combinatorial proof and the mathematical proof

Answer 1

First off, the range of your indices is wrong: it should be

$$ \sum_{k=-q}^l \binom{l-k}{m} \binom{q+k}{n} $$

Indeed, I would write $t=l+q$ so that the identity can be written simply as

$$ \sum_{r+s=t}\binom{r}{m}\binom{s}{n}=\binom{t+1}{m+n+1}. $$

This is a generalized hockey-stick identity. (The usual HS identity is the special case where one of $m$ or $n$ is set to $0$.) A combinatorial argument proceeds as follows:

The RHS counts how many ways there are to choose $m+n+1$ elements out of the numbers $\{1,\cdots,t+1\}$. The special thing about the "$m+n+1$" is that the numbers you picked can be categorized into three types: the smallest $m$ numbers you picked, the largest $n$ numbers you picked, and one number you picked in between the other two categories, the $(m+1)$st largest number.

The $(m+1)$st largest number you picked not only separates the numbers you picked into the $m$ smallest on one side and $n$ largest on the other, it also separates all of the numbers $1,\cdots,t+1$ into $r$ smaller numbers on the left and $s$ larger numbers on the right, for some pair $r,s$ with $r+s=t$. Note this means the $(m+1)$st largest number you picked is $(r+1)$.

In other words, the process of picking $m+n+1$ numbers out of $1,\cdots,t+1$ can be split into the following process: first pick the $(m+1)$st largest number $(r+1)$, then pick $m$ smaller numbers out of $1,\cdots,r$ and $n$ larger numbers out of the last $s$ numbers from $1,\cdots,t+1$.

Here is a generating function proof. By differentiating the infinite geometric series formula $m$ times and then multiplying by $(-x)^m/m!$ we arrive at the generating function:

$$ \frac{x^m}{(1-x)^{m+1}}=\sum_{r=0}^\infty\binom{r}{m}x^r. $$

Then the generalized HS identity follows from equating coefficients of $x^t$ below:

$$ \frac{x^m}{(1-x)^{m+1}}\cdot\frac{x^n}{(1-x)^{n+1}}=\frac{1}{x}\frac{x^{m+n+1}}{(1-x)^{(m+n+1)+1}}. $$

(Compare with the generating function proof of the usual Vandermonde identity.)

Answer 2

Here we show OPs identity is a Chu-Vandermonde identity in disguise. The following is valid \begin{align*} \color{blue}{\sum_{k=-q+n}^{l-m}\binom{l-k}{m}\binom{q+k}{n}=\binom{l+q+1}{m+n+1}\qquad l,m,n,q\in\mathbb{N}}\tag{1} \end{align*} where we set lower and upper limit of the sum, so that the upper index of the binomial coefficients of the summands are greater or equal the lower index, since otherwise the binomial coefficients are zero.

We obtain \begin{align*} \color{blue}{\sum_{k=-q+n}^{l-m}}&\color{blue}{\binom{l-k}{m}\binom{q+k}{n}}\\ &=\sum_{k=0}^{l-m+q-n}\binom{l-k+q-n}{m}\binom{k+n}{n}\tag{2}\\ &=\sum_{k=0}^{l-m+q-n}\binom{l-k+q-n}{l-k+q-n-m}\binom{k+n}{k}\tag{3}\\ &=\sum_{k=0}^{l-m+q-n}\binom{-m-1}{l-m+q-n-k}\binom{-n-1}{k}(-1)^{l+q-m-n}\tag{4}\\ &=(-1)^{l+q-m-n}\binom{-m-n-2}{l-m+q-n}\tag{5}\\ &=\binom{l+q+1}{l-m+q-n}\tag{6}\\ &\,\,\color{blue}{=\binom{l+q+1}{m+n+1}}\tag{7} \end{align*} and the claim (1) follows.

Comment:

• In (2) we shift the index to start with $k=0$ and compensate this by substituting in the sum $k\to k+q-n$.

• In (3) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$ twice.

• In (4) we use the binomial identity $\binom{p}{q}=\binom{-p+q-1}{q}(-1)^q$ twice.

• In (5) we apply the Chu-Vandermonde identity.

• In (6) we apply again $\binom{p}{q}=\binom{-p+q-1}{q}(-1)^q$.

• In (7) we apply again $\binom{p}{q}=\binom{p}{p-q}$.