<s>[INST] Is there a binomial identity for this expression $\frac{\binom{r}{k}}{\binom{n}{k}}$? What I'm trying to prove is this summation:
$$\sum_{i=0}^{k} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k - i}}{\dbinom{n}{k}} \cdot i = \dfrac{r}{n} \cdot k$$
I used induction on $k$ as follows:
$$LHS = \sum_{i=0}^{k + 1} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k + 1 - i}}{\dbinom{n}{k + 1}} \cdot i$$
            $$ = \sum_{i=0}^{k} \dfrac{\dbinom{r}{i} \cdot \dbinom{n - r}{k - i}}{\dbinom{n}{k}} \cdot i 
   + \dfrac{\dbinom{r}{k + 1} \cdot \dbinom{n - r}{k + 1 - (k + 1)}}{\dbinom{n}{k+1}} \cdot (k + 1)$$
            $$ = \dfrac{rk}{n} + \dfrac{\dbinom{r}{k + 1} \cdot \dbinom{n - r}{0}}{\dbinom{n}{k+1}} \cdot (k + 1)$$
            $$ = \dfrac{rk}{n} + \dfrac{\dbinom{r}{k + 1}}{\dbinom{n}{k+1}} \cdot (k + 1)$$
Then I was stuck with the expression $\dfrac{\dbinom{r}{k + 1}}{\dbinom{n}{k + 1}}$, where $k + 1 \leq r \leq n$. I guess I need a clever trick here to simplify the left hand side, any idea?
 [/INST] We don't need induction: 
\begin{align*}
\sum_{i=0}^k\binom ri\binom{n-r}{k-i}i&=\sum_{i=1}^k\frac{r!}{(i-1)!(r-1-(i-1))}\binom{n-r}{k-i}\\
&=r\sum_{l=0}^{k-1}\binom{r-1}{l}\binom{n-r}{k-1-l}\\
&=r\binom{n-r+r-1}{k-1}\\
&=r\binom{n-1}{k-1}\\
&=\frac rn\frac{n!}{(k-1)!(n-k)!}\\
&=\frac {rk}n\binom nk,
\end{align*}
the third equality is Chu-Vandermonde identity.
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