<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] 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
$$\sum_{r=2}^9\binom{r}2\binom{12-r}3\binom{13}6\;.$$
</s>