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? 

Consider the number of non-negative integral solutions to $X_1 + X_2 + X_3 .. + X_7 = 7$. This should equal $\binom{7+ 7 - 1}{7 - 1} = \binom{13}{6}$.
These solutions can also be counted as follows:
Case 1: Let $X_1 + X_2 + X_3 = 0$ and $X_4 + X_5 + X_6 + X_7 = 7$. The combined number of solutions for these are $\binom{2}{2} \cdot \binom{10}{3}$
Case 2: Let Let $X_1 + X_2 + X_3 = 1$ and $X_4 + X_5 + X_6 + X_7 = 6$. The combined number of solutions for these are $\binom{3}{2} \cdot \binom{9}{3}$.
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Case 8: Let Let $X_1 + X_2 + X_3 = 7$ and $X_4 + X_5 + X_6 + X_7 = 0$. The combined number of solutions for these are $\binom{9}{2} \cdot \binom{3}{3}$.
Thus, we get that $$\binom{13}{6} = \binom{2}{2} \cdot \binom{10}{3} + \binom{3}{2} \cdot \binom{9}{3} + \ .. \ + \binom{9}{2} \cdot \binom{3}{3}$$ 
