How do you evaluate this sum of multiplied binomial coefficients: $\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} $? We have to find the value of x+y in:
$$\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} = \binom{x}{y} $$
My approach:
I figured that the required summation is nothing but the coefficient of $x^3$ is the following expression:
$$\sum_{r=1}^7 \frac{r(r+1)}{2}(1+x)^{10-r} $$
which looks like:
$$(1+x)^{10} + 3(1+x)^9 + 6(1+x)^8 + ..... + 36(1+x)^3$$
Hence we have a sequence in which the difference of the coefficients forms an AP.
So let the expression = $S$
So we have:
$$\frac {S}{1+x} = (1+x)^{9} + 3(1+x)^8 + 6(1+x)^7 + ..... + 36(1+x)^2$$
Subtracting $\frac{S}{1+x}$ from $S$, we get:
$$S(1-\frac{1}{1+x}) = (1+x)^{10} + 2(1+x)^9 + 3(1+x)^8 + ..... + 8(1+x)^3 - 36(1+x)^2$$
Here, all the terms except the last one are in AGP and hence we can find $S$ by applying the same method and isolating $S$ on one side. Then we can process to find the coefficient of $x^3$ but it got really lengthy. In the test that it was given, we have an average time of 3 minutes per question and not even the hardest question takes more than 7-8 minutes if you know how to do it. 
Hence, I'm wondering if there is a better, shorter way to solve this question. 
 A: $$\begin{align}\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} 
&=\sum_{r=2}^9\binom r{r-2}\binom {12-r}{9-r}\\
&=\sum_{r=2}^9(-1)^{r-2}\binom{-3}{r-2}(-1)^{9-r}\binom {-4}{9-r}&&\text{(upper negation)}\\\
&=-\sum_{r=2}^9\binom {-3}{r-2}\binom{-4}{9-r}\\
&=-\binom {-7}7&&\text{(Vandermonde)}\\
&=-(-1)^7\binom {13}7&&\text{(upper negation)}\\
&=\binom {13}7=\binom{13}6\quad\blacksquare
\end{align}$$
A: You can also do it with almost no computation if you make the right combinatorial argument. Let’s count the ways to choose $6$ numbers from the set $S=\{0,1,\ldots,12\}$. Let $r$ be the third-smallest of the $6$; how many ways are there to choose the other $5$ numbers?
There are $r$ members of $S$ less than $r$, so there are $\binom{r}2$ ways to choose the $2$ numbers smaller than $r$. Similarly, there are $12-r$ members of $S$ greater than $r$, so there are $\binom{12-r}3$ ways to choose the $3$ numbers larger than $r$. These choices are independent of each other, so there are $\binom{r}2\binom{12-r}3$ ways to choose the $6$ numbers so that the third-smallest is $r$.
Now sum over the possible values of $r$ to find that there are $\sum_{r=2}^9\binom{r}2\binom{12-r}3$ $6$-element subsets of $S$. Of course $|S|=13$, so there are $\binom{13}6$ such subsets, and we have
$$\sum_{r=2}^9\binom{r}2\binom{12-r}3\binom{13}6\;.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{r = 2}^{9}{r \choose 2}{12 - r \choose 3}} =
\sum_{r = 0}^{\infty}{r + 2\choose 2}{10 - r \choose 3} =
\sum_{r = 0}^{\infty}{r + 2\choose r}{10 - r \choose 7 - r}
\\[3mm] = &\
\sum_{r = 0}^{\infty}{-r - 2 + r - 1 \choose r}\pars{-1}^{r}{-10 + r + 7 - r - 1 \choose 7 - r}\pars{-1}^{7 - r}
\\[5mm] = &\
-\sum_{r = 0}^{\infty}{-3 \choose r}{-4 \choose 7 - r} =
-\sum_{r = 0}^{\infty}{-3 \choose r}\ \overbrace{%
\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-4} \over z^{8 - r}}
\,{\dd z \over 2\pi\ic}}^{\ds{{-4 \choose 7 - r}}}
\\[5mm] = &\
-\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-4} \over z^{8}}\
\overbrace{\sum_{r = 0}^{\infty}{-3 \choose r}z^{r}}^{\ds{\pars{1 + z}^{-3}}}\
\,{\dd z \over 2\pi\ic} =
-\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-7} \over z^{8}}\,{\dd z \over 2\pi\ic}
\\[5mm] = &\
-{-7 \choose 7} = -{-\pars{-7} + 7 - 1 \choose 7}\pars{-1}^{7} =
\color{#f00}{{13 \choose 7}} = 1716
\end{align}
