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?
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 $$\sum_{r=2}^n\binom{r}{n}\binom{n}{r}=2^{n-3} (n-1) n$$