# Combinatorial interpretation of a sum identity: $\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}$

I solved $\sum_{k=1}^n(k-1)(n-k)$ algebraically \begin{eqnarray*} \sum_{k=1}^n(k-1)(n-k)&=&\sum_{k=1}^n(nk-n-k^2+k)\\ &=&\sum_{k=1}^nnk-\sum_{k=1}^nn-\sum_{k=1}^nk^2+\sum_{k=1}^nk\\ &=&\frac{n(n^2+n)}{2}-n^2-\frac{n(2n^2+3n+1)}{6}+\frac{n^2+n}{2}\\ &=&n\left(\frac{3n^2+3n-6n-2n^2-3n-1+3n+3}{6}\right)\\ &=&n\left(\frac{n^2-3n+2}{6}\right)=\frac{n(n-1)(n-2)}{6}\\ &=&\frac{n!}{(n-3)!3!}=\binom{n}{3} \end{eqnarray*} But now I am interested in a combinatorial interpretation of it. For the right hand side $\binom{n}{3}$ is the number of ways to choose 3 from a total of $n$, and for the left hand side, it looks like dividing $n$ into 2 groups, one of size $k$, and other of size $(n-k)$, then sum over all possible $k$, but I do not see any clues that how I can choose 3 from these 2 groups.

• See also: math.stackexchange.com/questions/1113556/… (Basically the same question - up to a change from $n$ to $n+1$ - but asking for any proof, not only combinatorial proof.) Commented Jan 21, 2015 at 15:14

## 1 Answer

The left hand side represents picking a "middle" element in a set of $3.\$ Then you have $k-1$ choices for picking the smallest element and $n-k$ choices for picking the largest element.

For example if $n=5$ and $k=3$, then you have $2 \times 2$ ways of $3$ being the middle element:

$$(1 2) 3 (4 5)$$

If $n = 7$ and $k=5$, you have $4 \times 2$ ways for $5$ to be the middle element:

$$(1 2 3 4) 5 (6 7)$$

• Really sneaky. Commented Mar 26, 2013 at 18:16