the number of inversions in the permutation "reverse" Known, that number of inversions is $k$ in permutation: 
$$\begin{pmatrix}
 1&  ...&  n& \\ 
 a_1&  ...&  a_n& 
\end{pmatrix}$$
Find number of inversions in permutation (let's call it "reverse"): 
$$\begin{pmatrix}
 1&  ...&  n& \\ 
 a_n&  ...&  a_1& 
\end{pmatrix}
$$
First, max number of inversions is $\frac{n(n-1)}{2}$. Then we can see if we get permutation with $0$ inversions than "reverse" permutation for it will have $\frac{n(n-1)}{2}$. 
I can guess that for $\sigma(i)$, $\sigma(n-i+1)$ sum of inversions in them is $\frac{n(n-1)}{2}$. Thus, answer will be $\frac{n(n-1)}{2}-k$
Please help me to prove that.
 A: Supposing you define an inversion of $a$ as a pair of indices $i<j$ such that $a_i>a_j$, you can easily show that $(i,j)$ with $i<j$ is an inversion of $a$ if and only if $(n+1-j,n+1-i)$ is not an inversion of the reverse of $a$. The pair of values at those indices are the same in both cases, so if you define an inversion to be the (inverted) pair of values $\{a_i,a_j\}$ then it is even easier: the set of inversions of the reverse of $a$ is the complement of the set of inversions of$~a$.
A: Hints.


*

*Think of some permutation $ι$ on $[n]$ so that for any permutation $σ$ on $[n]$, “$σ$-reverse” is $σι$.

*An inversion of any permutation $σ$ on $[n]$ is a pair $(i,j)$ with $i < j$ such that $\frac{σ(i) - σ(j)}{i-j} < 0$.

*For permutations $σ$, $τ$ on $[n]$ you have $\frac{(στ)(i) - (στ)(j)}{i - j} = \frac{σ(τ(i)) - σ(τ(j))}{τ(i) - τ(j)}·\frac{τ(i) - τ(j)}{i - j}$.

*What does this imply for the inversions of $στ$ in terms of the inversions of $σ$ and $τ$?

*How many inversions does $ι$ have?

