VANDERMOND’S IDENTITY:
Let m, n, and r be nonnegative integers with r not exceeding either m or n. Then
$${m+n \choose r} = \sum_{k=0}^{r}{ {{m}\choose{r-k}} { {n} \choose {k} } }$$
The exercise asks to determine, using Vandermonde's identity, the value of n that makes the following true equality:
$${n+1 \choose 4} = {n \choose 3}$$
I believe that, since r = 4 > 1 = m, the formula can not be used. Am I right or is there some "generalized Vandermonde's identity" that validates the use of the formula in this case?
The Vandermonde's identity holds if $m,n,r$ are all non-negative integers. (Without the restriction $r$ is not exceeding either $m$ or $n$.) In particular, the identity holds for $r = 4$ and $m=1$.
Note that for two integers $a,b \geq 0$ with $a<b$, we define $${a \choose b} = 0,$$ since it is impossible to choose $b$ people from $a$ people.