Vandermonde's identity? How to continue? I have: 
$$\sum\limits_{k = 1}^{10}k\binom{10}{k}\binom{20}{10-k} = $$
and I know that it doesn't matter if $k = 0$ so it also equals:
$$= \sum\limits_{k = 0}^{10}k\binom{10}{k}\binom{20}{10-k} = $$
and now it really reminds Vandermonde's identity so it's tempting to write:
$$= \binom{30}{10} \cdot \sum\limits_{k = 1}^{10}k = \binom{30}{10} \cdot 10!$$
but it seems wrong because on smaller numbers the equations doesn't hold... What's the right way to continue in order to get an equivalent expression without sigma?
 A: Ok I found an algebric answer!
$$\begin{align}
&\sum\limits_{k=1}^{10}k\binom{10}{k}\binom{20}{10-k}\\= &\sum\limits_{k=0}^{10}k\binom{10}{k}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}k\cdot\frac{10!}{(k)!(10-k)!}\binom{20}{10-k} \\= &\sum\limits_{k = 0}^{10}\frac{10!}{(k-1)!(10-k)!}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}10\cdot \frac{9!}{(k-1)!(10-k)!}\binom{20}{10-k} \\ = &\sum\limits_{k = 0}^{10}10\cdot \frac{9!}{(k-1)!(9 - (k-1))!}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}10\cdot \binom{9}{k-1}\binom{20}{10-k} \\ = &10\cdot \sum\limits_{k = 0}^{10}\binom{9}{k-1}\binom{20}{10-k} \\= &10\cdot \sum\limits_{k = 0}^{9}\binom{9}{k}\binom{20}{9-k} \\= &10 \cdot \binom{29}{9}
\end{align}$$
A: Hint: Use Vandermonde identity:
http://en.wikipedia.org/wiki/Vandermonde%27s_identity
I suppose you can write it as
$S =\sum_{k=0}^{r} k\pmatrix{m \\ k}\pmatrix{n \\ r-k}$.
No put r = k-1 and write out the expression which is S'
$ S = r{(m+n-1)\choose r-1}$
Here in this case r = 10, m= 10, and n = 20.
$S = \sum_{k=0}^{10} k \pmatrix{10 \\ k}\pmatrix{20 \\ 10-k} = 10{29\choose9}$ 
A: The following does not answer the question, but does address finding a nice expression for the sum. 
A box has $10$ chocolate doughnuts and $20$ plain doughnuts. We grab $10$ doughnuts at random. Let random variable $X$ be the number of chocolate doughnuts we grab. Then 
$$\Pr(X=k)=\frac{\binom{10}{k}\binom{20}{10-k}}{\binom{30}{10}},$$
and therefore the sum of the problem is equal to $\dbinom{30}{10}E(X)$.
It remains to calculate $E(X)$. Let indicator random variable $Y_i$ be defined by $Y_i=1$ if the $i$-th doughnut we grab is chocolate, and $Y_i=0$ otherwise.
Then $X=Y_1+\cdots+Y_{10}$, so by the linearity of expectation we have $E(X)=E(Y_1)+\cdots +E(Y_{10})$.
Easily $E(Y_i)=\frac{10}{30}$, so $E(X)=10\cdot \frac{10}{30}$ and we are finished.
Remark: The argument took a detour through probability, specifically the mean. Perhaps one could call it a mean proof.
