Combinatoric Proof of $i(n,k) = i(n, \binom{n}{2}-k)$ Let $i(n,k)$ denote the number of permutation where $\sigma \in S_n$ with $i(\sigma) = k$, that is $k$ inversions.
I want to provide combinatoric proof for the below statement:
$i(n,k) = i(n, \binom{n}{2}-k)$
Any hint/advice to approach?
 A: Let's illustrate with an example:
$$\begin{array}{cc}&\\\text{Number}&\begin{array}{c|ccccc}
5&\bullet&&&&\\
4&&&&&\bullet\\
3&&&&\bullet&\\
2&&&\bullet&&\\
1&&\bullet&&&\\
 \hline&1&2&3&4&5\\
\end{array}\\&\text{Position}\end{array}$$
This represents the permutation:
$$5, 1, 2, 3, 4$$ 
and has the $4$ inversion pairs $(5,1)$, $(5,2)$, $(5,3)$ and $(5,4)$. On the graph these are pairs of points that decrease in height from left to right. 
Now if we turn this permutation graph upside-down:
$$\begin{array}{cc}&\\\text{Number}&\begin{array}{c|ccccc}
5&&\bullet&&&\\
4&&&\bullet&&\\
3&&&&\bullet&\\
2&&&&&\bullet\\
1&\bullet&&&&\\
 \hline&1&2&3&4&5\\
\end{array}\\&\text{Position}\end{array}$$
This is the permutation:
$$1,5,4,3,2$$
The previous $4$ pairs that were inversions become $(1,5)$, $(1,4)$, $(1,3)$, $(1,2)$ i.e they are now the only pairs that are not inversions. 
So since there are $\binom{5}{2}$ pairs of points in total, every permutation of $[5]$ with $4$ inversions is in bijection with every permutation with $\binom{5}{4}-4$ inversions simply by inverting it's permutation graph. Hence:
$$i(5,4)=i(5,\tbinom{5}{2}-4)$$
This argument clearly holds in general so:
$$i(n,k)=i(n,\tbinom{n}{2}-k)$$
