<s>[INST] Simplify $\sum^{20}_{k=10} k\binom{k-1}{9}$. 
Simplify $$\sum^{20}_{k=10} k\binom{k-1}{9}$$ as much as possible.

I feel like I could utilize the hockey stick identity, but have not found a way to do so with that extra $k$.
Any help would be appreciated.
 [/INST] If you couldn't think about how to get rid of the '$k$':
What we need to evaluate is
\begin{align*}
\sum^{20}_{k=10} k\binom{k-1}{9}&=20{19\choose 10}+19{18\choose 9}+\cdots+10{9\choose 9}\\
\end{align*}
Using Hockey-stick identity, we can evaluate
\begin{align*}
\sum^{20}_{k=10} k\binom{k-1}{9}&=10{9\choose 9}+\cdots+2{17\choose 9}+{18\choose 10}\\
\sum^{20}_{k=10} k\binom{k-1}{9}&=\underbrace{\underbrace{{9\choose 9}+{10\choose 9}+\cdots+{18\choose 9}}_{{19\choose 10}+}+\underbrace{{9\choose 9}+{10\choose 9}+\cdots+{17\choose 9}+\cdots}_{{18\choose 10}+\cdots}+\underbrace{{9\choose 9}}_{{10\choose 10}}}_{\color{red}{20\choose 11}}\\
\end{align*}
But we seek $20\displaystyle\sum^{20}_{k=10} \binom{k-1}{9}-\color{red}{20\choose 11}=20{20\choose10}-{20\choose 11}=10{21\choose 11}$.
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