Summation of hypergeometric probability multiplied by some factor

Question

I am not sure how to prove the following: $$\sum _{s=0}^{k}{\frac {{j\choose k-s}{n-j\choose s}s}{{n\choose k}}}=-{\frac { \left( -n+j \right) k}{n}}$$

I know if the factor $s$ is omitted, the sum just becomes 1 because of Vandermonde's identity (I think!).

I am not a mathematician and so this is super confusing!

Answer 1

First note that the binomial coefficient in the denominator is not dependent on the summation variable & can be factored out. So we only need to consider \begin{eqnarray*} \sum_{s=0}^{k} s \binom{j}{k-s} \binom{n-j}{s}= (n-j) \sum_{s=0}^{k} \binom{j}{k-s} \binom{n-j-1}{s-1}. \end{eqnarray*} This is just Vandermonde again the result you require follows easily.

Answer 2

Hint: $$\binom{n-j}ss=\binom{n-j-1}{s-1}(n-j)$$ Factor the $n-j$ out of the summation, then apply Vandermonde's identity.