Number of inversions Compute the sum of the number of inversions that appear in the elements of $S_n$. In other words find the total number of inversions that the elements of $S_n$ have combined.
I mean how can we calculate the sum of the inversions if we don't know about the elements present in $S_n$?
 A: Observe that an element $k$  of a permutation of $S_n$ can participate
in  zero,  one, two  etc.  up to  $k-1$  inversions.  Hence we obtain  the
following  generating  function of  permutations  of $S_n$  classified
according to inversions:
$$G(z) = 1\times (1+z)\times (1+z+z^2)\times\cdots\times
(1+z+z^2+\cdots+z^{n-1}).$$
This is
$$G(z) = \prod_{q=0}^{n-1} (1+z+z^2+\cdots +z^q).$$
The total number of inversions is thus given by
$$\left.\frac{d}{dz} G(z)\right|_{z=1}
= \left.\prod_{q=0}^{n-1} (1+z+z^2+\cdots +z^q)
\sum_{q=0}^{n-1} 
\frac{1+2z+3z^{2}+\cdots+qz^{q-1}}{1+z+z^2+\cdots +z^q}\right|_{z=1}
\\ = n! \sum_{q=0}^{n-1} \frac{1/2 q(q+1)}{q+1}
= \frac{1}{2} n! \sum_{q=0}^{n-1} q
= \frac{1}{2} n! \frac{1}{2} (n-1) n 
= \frac{1}{4} n! (n-1) n.$$
This yields for the average  number  of inversions in a random permutation
$$\frac{1}{4} (n-1) n.$$
The generating function $G(z)$ also appeared at this MSE link.
A: Shorter answer: 
Given a permutation $\pi\in S_n$, let $X_\pi (i,j)=1$ if $\pi(i)>\pi(j)$ and $0$ otherwise. (That is, $X$ tells whether $i$ and $j$ are inverted in $\pi$.)
Then the number of inversions of $\pi$ is $i_\pi = \displaystyle \sum_{(i,j)} X_\pi (i,j)$, where the sum -- and similar sums below -- are over all $(i,j)$ such that $i,j\in \{1,2,\ldots,n\}$ and $i<j$. The expected value of $i_\pi$ is
$${1\over n!}\sum_{\pi} i_\pi = {1\over n!} \sum_\pi \sum_{(i,j)} X_\pi (i,j)
= {1\over n!} \sum_{(i,j)} \sum_\pi X_\pi(i,j) = {1\over n!} \sum_{(i,j)} {n!\over 2}= {1\over2}\sum_{(i,j)} 1 = {1\over 2}\cdot {n \choose 2}.
$$
Notes: 
$\displaystyle\sum_\pi X_\pi(i,j)={n! \over 2}$, because $i$ and $j$ are inverted in exactly half of the permutations.
$\displaystyle\sum_{(i,j)}$ is over exactly $\displaystyle{n\choose 2} = {n(n-1)\over 2}$  ordered pairs.
A: Shorter answer (:-)
For each permutation $\sigma\in\mathfrak{S}_n$, let's denote by $\text{inv}(\sigma)$ the number of its inversions.
Consider the permutation $c$ defined by :
$$\forall k\in\{1,\ldots,n\},\,c(k)=n+1-k$$
Since $\mathfrak{S}_n\to\mathfrak{S}_n,\sigma\mapsto\sigma\circ c$ is bijective, and since any pair $\{i,j\}$ is an inversion for $\sigma$ iff it's NOT an inversion for $\sigma\circ c$, we have :
$$2\sum_{\sigma\in\mathfrak{S}_n}\text{inv}(\sigma)=\sum_{\sigma\in\mathfrak{S}_n}\left[\text{inv}(\sigma)+\text{inv}(\sigma\circ c)\right]=n!\frac{n(n-1)}2$$ and finally :
$$\boxed{\sum_{\sigma\in\mathfrak{S}_n}\text{inv}(\sigma)=n!\frac{n(n-1)}4}$$
Remark - This proves that the expected value of the number X of inversions of a randomly choosen permutation is : $\mathbb{E}(X)=\frac{n(n-1}4$
