Show that $\sum_{k=0}^{\infty} \binom{r}{k} \binom{s}{n+k} = \binom{r+s}{r+n}$.

Question

I can't resolve this exercise and I need some tips.

Let be $n$ a integer, $s$ a real number and $r \geq 0$ a integer. Show that

$$ \sum_{k=0}^{\infty} \binom{r}{k} \binom{s}{n+k} = \binom{r+s}{r+n} $$

Answer 1

Suppose we seek to evaluate $$\sum_{k=0}^r {r\choose k} {s\choose n+k}.$$

Start from $${s\choose n+k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+k+1}} (1+z)^s \; dz.$$

This gives the following integral for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \sum_{k=0}^r {r\choose k} \frac{1}{z^{n+k+1}} (1+z)^s \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^s}{z^{n+1}} \sum_{k=0}^r {r\choose k} \frac{1}{z^k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^s}{z^{n+1}} \left(1+\frac{1}{z}\right)^r \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^s}{z^{n+1}} \frac{(1+z)^r}{z^r} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^{r+s}}{z^{r+n+1}} \; dz.$$

This last integral can be evaluated by inspection and is given by $$[z^{r+n}] (1+z)^{r+s} = {r+s\choose r+n}.$$ This obviously confirms the combinatorial proof, which is simple.

A trace as to when this method appeared on MSE and by whom starts at this MSE link.

Answer 2

Both sides are polynomials in $s$ (the terms of the series are zero when $k>r$), so it suffices to show they are equal when $s$ is a nonnegative integer.

Suppose that I have $s$ statesmen and $r$ robots. ${{s+r}\choose{n+r}}$ counts the number of ways that we can form a caucus with $n+r$ members.

But if our caucus has $n+k$ statesmen ($0\leq k \leq r$), then we must exclude exactly $k$ robots. So the number of such caucuses is ${r\choose k}{s \choose {n+k}}$.

Answer 3

Hint: $\binom{r}{k}=\binom{r}{r-k}$. How many ways can you choose $r+n$ persons out of $r$ men and $s$ women?