<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] Numerical method. 
The sum is equal to $1716$.
Now we want to find $x,y$ in:
$$\frac{x!}{y!(x-y)!}=1716=2^2\cdot 3\cdot 11\cdot 13=\frac{2^3\cdot 3\cdot 11\cdot 13}{2!}=\frac{2^3\cdot 9\cdot 11\cdot 13}{3!}=\frac{2^5\cdot 9\cdot 11\cdot 13}{4!}=$$
$$\frac{2^4\cdot 9\cdot 10\cdot 11\cdot 13}{5!}=\frac{8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13}{6!}=\frac{13!}{6!\cdot 7!}=\frac{13!}{6!\cdot (13-6)!}.$$
Hence: $x+y=13+6=19$.
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