# Determining the closed form of $\sum\limits_{r = 2}^n \binom{n}{r} \binom{r}{2}$ [closed]

Here is a sum in combinatorics, $$\sum\limits_{r = 2}^n \binom{n}{r} \binom{r}{2}$$ where $$n>2$$, does this have a closed form?

Yes, the closed form is

$$\sum\limits_{r = 2}^n {C_n^rC_r^2} = n \cdot (n-1) \cdot 2^{n-3}$$

Seems there're various ways to prove this.

In how many ways can we choose from $$n$$ people a committee with $$r$$ members (where $$2 \le r \le n$$), and also choose $$2$$ chairmen from the committee. Express this number in two different ways and you get the above identity.
$$\sum_{r=2}^n \binom{n}{r} \binom{r}{2} =\sum_{r=2}^n \binom{n}{2} \binom{n-2}{r-2} =\binom{n}{2} \sum_{r=2}^n \binom{n-2}{r-2} =\binom{n}{2} 2^{n-2} =n(n-1) 2^{n-3}$$
We have the ordinary generating function for $$(1+x)^n = \sum_{0\leq r\leq n }{ n\choose r} x^{ r}\cdots (1)$$ Now we differentiate $$(1)$$ twice and then divide by $$2!$$ giving us $$\frac{(n)_2}{2!}(1+x)^{n-2}=\sum_{0\leq r\leq n} {n\choose r}\frac{(r)_2}{2!} x^{n-2}\\ {n\choose 2} (1+x)^{n-2}=\sum_{0\leq r\leq n} {n\choose r}{r\choose 2} x^{n-2}\cdots (2)$$ and where $$\displaystyle \frac{(m)_n}{n!}={m\choose n}$$ and setting $$x=1$$ in $$(2)$$ yield $$\sum_{0\leq r\leq n}{n\choose r}{r\choose 2} =2^{n-2}{n\choose 2}=2^{n-2}T_n= 2^{n-3}n(n-1)$$ note that for $$0\leq r\leq 2$$ the coefficients $${r\choose 2} =0$$ and hence we have $$\sum_{r=2}^{n}{n\choose r}{r\choose 2} = 2^{n-3}n(n-1)$$