<s>[INST] 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. 
 [/INST] $$\begin{align}\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} 
&=\sum_{r=2}^9\binom r{r-2}\binom {12-r}{9-r}\\
&=\sum_{r=2}^9(-1)^{r-2}\binom{-3}{r-2}(-1)^{9-r}\binom {-4}{9-r}&&\text{(upper negation)}\\\
&=-\sum_{r=2}^9\binom {-3}{r-2}\binom{-4}{9-r}\\
&=-\binom {-7}7&&\text{(Vandermonde)}\\
&=-(-1)^7\binom {13}7&&\text{(upper negation)}\\
&=\binom {13}7=\binom{13}6\quad\blacksquare
\end{align}$$
</s>