Simplify $\sum_{k=0}^n (k*(k-1)*{n \choose n-k})$ $$\sum_{k=0}^n (k*(k-1)*{n \choose n-k})$$
I an trying to simplify this as much as possible for $n = 24$.
I don't know if what I'm doing is right:
First of all I replaced ${n \choose n-k}$ with ${n \choose k}$
Then I distributed the sigmas like this:
$$\sum_{k=0}^n (k*(k-1)*{n \choose k}) = \sum_{k=0}^nk *\sum_{k=0}^n(k-1)*\sum_{k=0}^n{n \choose k}$$
Am I only allowed to do this in this step ?
 A: We have:
$$\sum_{k = 0}^{n}k(k-1)\binom{n}{n- k} = \sum_{k=0}^{n}k(k-1)\frac{n!}{k!(n -k)!} = \sum_{k=0}^{n}\frac{n!}{(k-2)!(n-k)!}$$
$$=\sum_{k=0}^{n}\frac{n(n-1)(n-2)!}{(k-2)!\big((n-2)-(k-2)\big)!} = n(n-1)\sum_{k=0}^{n}\binom{n-2}{k-2}$$
Discarding the first $2$ terms, which are $0$:
$$=n(n-1)\sum_{k-2 = 0}^{n-2}\binom{n-2}{k-2} = \boxed{n(n-1)2^{n-2}}$$
A: Recognize the sum you are given is the answer to the question of "Given a committee of size $n$, how many ways can we make a subcommittee of any size with a chairperson and a vice chairperson?"
It is the answer to that question by approaching the question in the following way:  First break apart based on the size of the subcommittee.  Let the size of the subcommittee be called $k$.  Then, pick who all isn't on the subcommittee at all (thereby picking who is on the subcommittee).  From those who weren't selected (and so are on the subcommittee) pick one of them to be the chairperson and then pick someone else to be the vicechair.
An alternate approach to counting can be done by from the original $n$ person committee first picking a person to be the chairperson for the subcommittee and then picking the vicechair from those left in $n(n-1)$ ways.  Now, for the remaining $n-2$ people, pick whether each was included or not included in the subcommittee.
This second approach yields an answer of:
$$n(n-1)2^{n-2}$$
By principles of double counting, since the original expression and this second expression both count the same scenario, they must be equal.  No messy algebraic manipulation needed.
