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? 
 A: 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 $$\binom{13}{6} = \binom{2}{2} \cdot \binom{10}{3} + \binom{3}{2} \cdot \binom{9}{3} + \ .. \ + \binom{9}{2} \cdot \binom{3}{3}$$ 
A: 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.
A: 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: $x+y=13+6=19$.
