<s>[INST] Finding value of $\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$ 
Finding value of $$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$

Solution I tried: 
$$(1+x)^{10}=1+\binom{10}{1}x+\binom{10}{2}x^2+\cdots +\binom{10}{10}x^{10}$$
$$(x+1)^{10}=x^{10}+\binom{10}{1}x^9+\binom{10}{2}x^7+\cdots +\binom{10}{10}$$
I did not find how do multiply terms such that i get my result. Help me 
 [/INST] $$S_{(10,2)} = \sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$
You can generalize the question as:
\begin{align}
S_{(n,\,r)} &= \sum^{n}_{k=r}\binom{k}{r}\binom{(n+r)-k}{r}\\
\end{align}

Set $(n+r) = m\,$ and we get 
\begin{align}
S_{(n,\,r)} &= \sum^{n}_{k=r}\binom{k}{r}\binom{m-k}{r}= \binom{m+1}{2r+1}\\
\end{align}
(See Another form of the Chu–Vandermonde identity and this proof )

So

$$S_{(n,\,r)} = \binom{n+r+1}{2r+1}$$

In your case
$$S_{(10,2)} = \binom{10+2+1}{2\cdot 2+1} = \binom{13}{5} = 1287$$
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