<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] It is Chu–Vandermonde identity
$$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}=\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{4-2}={12+1\choose 4+1}=1287.$$
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