# Value of expression $\sum\limits^{10}_{r=2}\binom{r}{2}\cdot \binom{10}{r}$

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?

$$\sum_{r=2}^{10}\binom{r}{2}\binom{10}{r}$$ 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 $$\sum_{r=2}^{10}\binom{r}{2}\binom{10}{r}=\binom{10}{2}\cdot2^8=\boxed{11520}.$$
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 $$\frac {10!} {8!2!} (1+1)^{8}$$.
Hint: It is $$\sum_{r=2}^n\binom{r}{n}\binom{n}{r}=2^{n-3} (n-1) n$$