Derive a Combinatorial Identity An urn exists with $x$ red balls and $y$ green balls. Draw $n$ balls without replacement. 
Derive a combinatorial identity for nonnegative integers $n$, $x$, and $y$, satisfying $1 \leq n \leq x + y$ from consideration of the following:
$$na = \sum_{k=0}^n k {n \choose k}a^k (1-a)^{n-k}$$
Attempt: I know that one side of the identity is $$n{x\over x+y}$$
Edit: The question is not to derive the na = ... identity above. It is to derive a new identity, using that identity as a guide, for drawing balls from the urn without replacement.
 A: \begin{aligned}
\sum^n_{k=0}k\binom{n}{k}a^k(1-a)^{n-k}&=\sum^n_{k=1}\frac{n(n-1)!}{(k-1)!(n-k)!}a^k(1-a)^{n-k}\\
&=na\sum^n_{k=1}\binom{n-1}{k-1}a^{k-1}(1-a)^{(n-1)-(k-1)}\\
&=na\sum^n_{k=0}\binom{n-1}{k}a^k(1-a)^{(n-1)-k}\\
&=na(a+1-a)^{n-1}
\end{aligned}
A: Oliver's answer is great, but just in case you like derivatives like me,
\begin{align}
\frac{\partial}{\partial y}(x+y)^n = \sum_{k=0}^n { n \choose k} x^{n-k} \frac{\partial}{\partial y}y^k\newline
n(x+y)^{n-1}=\sum_{k=0}^n { n \choose k} k x^{n-k}  y^{k-1}
\end{align}
Taking $x = (1-a)$ and $y = a$,  $x+y = 1$,
\begin{align}
na=\sum_{k=0}^n { n \choose k} k (1-a)^{n-k}  a^{k}
\end{align}
A: The expression provided as reference is the expected number of balls, having prob. $a$ of being extracted 
with replacement from an urn containing a total of $n$ balls.
$n x/(x+y)$ is the expected number of balls "type x" in $n$ balls extracted without replacement
from $x+y$ total balls: a sample mean.
You have $\binom{x}{k}$ ways to extract a $k$-subset from the set of $x$ balls (labelled)
and $\binom{y}{n-k}$ to extract $n-k$ from the $y$. In total
$$
\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,n} \right)} {\left( \matrix{
  x \cr 
  k \cr}  \right)\left( \matrix{
  y \cr 
  n - k \cr}  \right)}  = \left( \matrix{
  x + y \cr 
  n \cr}  \right)
$$
ways.
Then
$$
n{x \over {x + y}} = {1 \over {\left( \matrix{
  x + y \cr 
  n \cr}  \right)}}\sum\limits_{\left( {0\, \le } \right)k\,\left( { \le \,n} \right)} {k\left( \matrix{
  x \cr 
  k \cr}  \right)\left( \matrix{
  y \cr 
  n - k \cr}  \right)} 
$$
