Combinatorial interpretation of a sum identity: $\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}$
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:
$$(1 2 3 4) 5 (6 7)$$