Showing $\sum_{k = 2}^n \binom{k}{2} \binom{n}{k} = \binom{n}{2} 2^n$ without induction. How do I prove the identity$$\sum_{k = 2}^n \binom{k}{2} \binom{n}{k} = \binom{n}{2} 2^{n-2}$$combinatorially, i.e. counting the cardinality of the same set in two different ways? I know how to do it by induction, but I am at a loss on how to do it combinatorially. Thanks in advance.
 A: I think there's a mistake, and it should be $2^{n - 2}$ rather than $2^n$. For example, if $n = 2$ the left side is $1$ while the right side would be $2^2$.

You have a box containing $n$ items; you choose $k$ of them out of the box, and then choose $2$ out of these $k$. The end result is that you choose exactly $2$ items out of $n$. But our two items are selected as many times as there are sets containing those two items - this is equal to the number of subsets of the remaining $n - 2$ items - which is $2^{n - 2}$.
Hence, the sum is
$$\sum_{k = 2}^{n} \binom{n}{k} \binom{k}{2} = 2^{n - 2} \binom{n}{2}$$
A: Suppose you are interesting in count how many teams you can do with $n$ persons, which of the team having $2$ captains. On one side, it is the same that choose first the two captains, and then complete all the teams with each subset of the other $n-2$: $\binom{n}{2}2^{n-2}$
On the other hand, this is the same to count all the teams with 2 elements, 3 elements, 4 elements,..., $n$ elements and, for each $k\in\{2,...,n\}$ select the two captains from they: $\binom{n}{k}\binom{k}{2}$. Thus, the total is $\sum_{k=2}^n\binom{n}{k}\binom{k}{2}$
Since both procedements counts the same number of teams, we have $\sum_{k=2}^{n}\binom{n}{k}\binom{k}{2}=\binom{n}{2}2^{n-2}$
A: $$\binom k2\binom nk=\dfrac{k(k-1)}2\cdot\dfrac{n(n-1)\cdot(n-2)!}{\{n-2-(k-2)\}!\cdot k(k-1)\cdot(k-2)!}=n(n-1)\binom{n-2}{k-2}$$ for $k\ge2$
and $$\sum_{k=2}^n\binom{n-2}{k-2}=\sum_{r=0}^{n-2}\binom{n-2}r$$
can be found by setting $a=b=1$ in the binomial expansion of $$(a+b)^{n-2}$$
