prove the formula and then evalute the sum m,n,r are given non-negative integers, show that
$\sum_{k>=-n}$
${r \choose m+k}$
${s \choose n+k}$
$=$
${r+s \choose r-m+n}$
Then evaluate
$\sum_{k>=0}k$
${r \choose k}$
${s \choose k}$
I tried mathematical induction base on size of r. The base case is when r equals m,but then I found that 2 sides are not equal.
When r equals m,
$LHS = \sum_{k>=0}$
${m \choose m+k}$
${s \choose n+k}$
$=$
${m \choose m+0}$
${s \choose n+0}$
$={m \choose m}{s \choose n}=\frac{s(s-1)...(s-n+1)}{n!}$
$RHS$
$={m+s \choose n}=\frac{(s+m)(s+m-1)...(s+m-n+1)}{n!}$
 A: Actually, I agree on your calculation.
If the formula would be generally true, it should be true for r = m. Check carefully for typo's, and with this calculation go back to whoever gave you this problem.
(Not a real answer, but an opportunity for discussion, that shouldn't be hidden in the chat discussion.)
A: The first result is false as stated. Take $m=n=1$, $r=s=2$:
$$\binom2{1+0}^2+\binom2{1+1}^2=5\ne 6=\binom{2+2}{2-1+1}\;.$$
To see what it should be, note first that
$$\sum_{k\ge 0}\binom{r}{m+k}\binom{s}{n+k}=\sum_{k\ge 0}\binom{r}{r-m-k}\binom{s}{n+k}\;.$$
Let $\ell=k+n$, so that $k=\ell-n$; then
$$\binom{r}{r-m-k}\binom{s}{n+k}=\binom{r}{r-m+n-\ell}\binom{s}\ell\;,$$
and Vandermonde’s identity says that
$$\sum_{\ell=0}^{r-m+n}\binom{r}{r-m+n-\ell}\binom{s}\ell=\binom{r+s}{r-m+n}\;.$$
As $\ell$ runs from $0$ to $r-m+n$, $k$ runs from $-n$ to $r-m$. Thus, the correct statement of the first result is 
$$\sum_{k\ge -n}\binom{r}{m+k}\binom{s}{n+k}=\binom{r+s}{r-m+n}\;.$$
Since $\binom{s}{n+k}=0$ for $k<-n$, we can in fact simply write this
$$\sum_k\binom{r}{m+k}\binom{s}{n+k}=\binom{r+s}{r-m+n}\;.$$
(Vandermonde’s identity is proved at the link and at several places on this site, in case you’re not already familiar with it.)
For the second problem, note that $k\dbinom{r}k=r\dbinom{r-1}{k-1}$.
