<s>[INST] How to evaluate $\binom{2}{2}\binom{10}{3} + \binom{3}{2}\binom{9}{3} + \binom{4}{2}\binom{8}{3} + \ldots + \binom{9}{2}\binom{3}{3}$ I really can't understand how to approach exercise 41. 
I tried by comparing the 5th term of binomial expansion and something like that, but getting 12C5, I think the answer should be 13C6. How to approach? 
 [/INST] We wish to show that 
$$\sum_{k = 2}^{9} \binom{k}{2}\binom{10 - k}{3} = \binom{13}{6}$$
The right hand side counts six-element subsets of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\}$.
The left-hand side counts six-elements subsets of the set whose third-largest element is $k + 1$.  To see this, observe that if the third-largest element is $k + 1$, we must choose two of the $k$ elements smaller than $k + 1$ and three of the $13 - (k + 1) = 12 - k$ elements larger than $k + 1$.  Hence, the number of subsets of size six whose third-largest element is $k + 1$ is 
$$\binom{k}{2}\binom{10 - k}{3}$$
For $k + 1$ to be the third-largest element, $k$ must be at least $2$ and at most $9$ since there be must be at least two numbers smaller than $k + 1$ and at least three numbers larger than $k + 1$.  Since every six-element subset of the set has a third-largest element, the number of all six-element subsets of the set  is 
$$\sum_{k = 2}^{9} \binom{k}{2}\binom{10 - k}{3}$$
Hence, the identity holds.
</s>