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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?

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$\sum_{r=2}^{10}\binom{r}{2}\binom{10}{r}$ 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 $$\sum_{r=2}^{10}\binom{r}{2}\binom{10}{r}=\binom{10}{2}\cdot2^8=\boxed{11520}.$$

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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 $\frac {10!} {8!2!} (1+1)^{8}$.

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Hint: It is $$\sum_{r=2}^n\binom{r}{n}\binom{n}{r}=2^{n-3} (n-1) n$$

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