# Proof through combinatorial argument

I am attempting to solve this counting problem through combinatorial argument. The following is the equation I am given:

$$\sum_{i=1}^n (i-1)(n-i) = \binom{n}{3}$$

I understand that the right-hand side of this equation represents a set of $$n$$-elements out of which we choose 3. For example I believe we can say suppose we have a group of $$n$$ people and we want to choose 3 out of $$n$$ to be in a committee. However I'm not sure how to express the left-hand side in words. If forming a committee is an appropriate way to tackle this problem then I know the left-hand side must utilize the addition and multiplication principles, but I don't know how to put it into words. Also my intuition tells me that in solving this we should first flip $$(n-i)(i-1)$$

Thanks!

• flipping is not needed. See my answer. – Henno Brandsma Feb 21 '16 at 8:41
• "flipping" is not needed. But the flipped form certainly has more visual appeal. i spends its time between n and 1 so why not put it between n and 1? – candied_orange Feb 23 '16 at 3:36

Hint: split on the fact that the middle member (in sorted numerical order, the members are numbered $1$ to $n$) is $i$. Then we pick one from before and one from after.
• Then, doesn't $i$ go from $2&-$n-1$?. If$i$is either of the end points, then only one other number can be chosen. – user230452 Feb 21 '16 at 9:07 • We split the three person committees in disjoint sets$M_i$, which consists of all committees$a < b < c$($a,b,c \in \{1,\ldots,n\}$) with$b = i$. This is a partition of all committees, and so we have the sum. If we want to pick a committee in$M_i$, we are forced to take$b=i$but we need exactly one member form$\{1,\ldots i-1\}$and one from$\{i+1,\ldots,n\}\$. Hence the product (2 independent choices). – Henno Brandsma Feb 21 '16 at 9:49