The value of expression $\displaystyle \sum\limits^{10}_{r=2}\binom{r}{2}\cdot \binom{10}{r}=$
What I tried:
$$\sum^{10}_{r=2}\frac{r!}{2!\cdot (r-2)!}\times \frac{10!}{r!\cdot (10-r)!}$$
$$\frac{10!}{2!}\sum^{10}_{r=2}\frac{1}{(r-2)!\times (10-r)!}$$
How do I solve it?
To continue from where you left off you get $\frac {10!} {8!2!} \sum\limits_{r=2}^{10} \binom {8}{r-2}=\frac {10!} {8!2!} \sum\limits_{s=0}^{8} \binom {8}{s}$ which is $\frac {10!} {8!2!} (1+1)^{8}$.
$\sum_{r=2}^{10}\binom{r}{2}\binom{10}{r}$ counts the number of teams that can be built from a group of ten people, with two leaders assigned. We could also count this by choosing the leaders beforehand in one of $\binom{10}{2}$ ways, and then succesively choosing whether to add or not each of the next eight people in one of $2^8$ ways. Therefore, we have $$\sum_{r=2}^{10}\binom{r}{2}\binom{10}{r}=\binom{10}{2}\cdot2^8=\boxed{11520}.$$
Hint: It is $$\sum_{r=2}^n\binom{r}{n}\binom{n}{r}=2^{n-3} (n-1) n$$
