How to find (or 'generate') combinatorial meaning for the given expression $\left(\dfrac{6(k-n)(k-1)}{(n-2)(n-1)}+1\right)\dfrac{30}{n(n+1)(n+2)}$
(for $n\geq 3$ and $1\leq k \leq n$)
The expression comes from question https://math.stackexchange.com/questions/304876/please-help-to-find-function-for-given-inputs-and-outputs where it is used to get answer for some unknown problem (I believe that's combinatorial, with possibility of used Monte-Carlo method). The OP didn't give any interpretation of the expression.
Can you please share your best ideas about combinatorial interpretation of the expression? Big thanks in advance.
P.S. There's tiny possibility that it was mathematical/SE joke.

This post does not show a meaning of the expression, but explains how one might arrive at it.
I want a quadratic function with zero mean defined on $\{1,\dots,n\}$. Naturally, it should be symmetric about the midpoint of the interval. The obvious symmetric function is $k(k-n-1)$, but it does not have mean zero. The mean is 
$$\frac{1}{n}\sum_{k=1}^n k(k-n-1) = -\frac{(n+1)(n+2)}{6}$$
So I subtract that, arriving at $k(k-n-1)+\frac{(n+1)(n+2)}{6}$. But maybe I also care about the second moment, which is 
$$\sum_{k=1}^n \left(k(k-n-1)+\frac{(n+1)(n+2)}{6}\right)^2 = \frac{(n-2)(n-1)n(n+1)(n+2)}{180}$$
Dividing by the second moment yields
$$\frac{180}{(n-2)(n-1)n(n+1)(n+2)}\left(k(k-n-1)+\frac{(n+1)(n+2)}{6}\right) $$
which is the formula in question.
