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?

Ok I found an algebric answer!
$$\begin{align}
&\sum\limits_{k=1}^{10}k\binom{10}{k}\binom{20}{10-k}\\= &\sum\limits_{k=0}^{10}k\binom{10}{k}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}k\cdot\frac{10!}{(k)!(10-k)!}\binom{20}{10-k} \\= &\sum\limits_{k = 0}^{10}\frac{10!}{(k-1)!(10-k)!}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}10\cdot \frac{9!}{(k-1)!(10-k)!}\binom{20}{10-k} \\ = &\sum\limits_{k = 0}^{10}10\cdot \frac{9!}{(k-1)!(9 - (k-1))!}\binom{20}{10-k} \\=&\sum\limits_{k = 0}^{10}10\cdot \binom{9}{k-1}\binom{20}{10-k} \\ = &10\cdot \sum\limits_{k = 0}^{10}\binom{9}{k-1}\binom{20}{10-k} \\= &10\cdot \sum\limits_{k = 0}^{9}\binom{9}{k}\binom{20}{9-k} \\= &10 \cdot \binom{29}{9}
\end{align}$$
