How to find the number of $2$ inversions in $S_5$? How would I find the number of permutations in $S_5$ that have exactly $2$ inversions?
By inversion I mean if $i<j$ but $\sigma(i)>\sigma(j)$ i.e the order in which an element appears is switched in the permutation.
I have found the following $6$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\ 
1 & 2 & 5 & 3 &4 
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\ 
1 & 4 & 2 & 3 &5 
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\ 
3 & 1 & 2 & 4 &5 
\end{pmatrix}$$
simply by pulling the element $5$ back two places and then $4$ back two places and then the same for $3$. This no longer works with $2$ because we have to move it to $5$ and this gives more than two inversions however we can reverse it and move $1$ forward two places and then move $2$ forward two places and likewise for $3$ this gives.
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\ 
2 & 3 & 1 & 4 &5 
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\ 
1 & 3 & 4 & 2 &5 
\end{pmatrix}$$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\ 
1 & 2 & 4 & 5 &3 
\end{pmatrix}$$
this gives us $6$ permutations with only $2$ inversions.
Is this all? Can anyone show me a better way that just trying to figure them out?
Thanks!

The generating function for inversions is the factorial $q$-analogue 
$$\left[n\right]_q!$$
The  $q$-analogue of $k$ is 
$$\left[k\right]=1+q+q^2+q^3+\cdots +q^{k-1}$$
so the factorial $q$-analogue is
$$\left[n\right]_q!=1(1+q)(1+q+q^2)(1+q+q^2+q^3)\cdots (1+q+q^2+q^3+\cdots +q^{n-1})$$
The number of permutations with $k$ inversions is given by the $x^k$ coefficient $A_{n,k}$ of
$$\left[n\right]_q! = \sum_{k=0}^{\binom{n}{2}}A_{n,k}x^k$$
in your case you want $A_{5,2}$. 
There is a simple proof of this inversion generating function in Theorem 1.2 page 3 of this book
Multiplying these polynomials is either a job for a computer algebra system like sage or we can do it manually. 
In sage if we input
show(expand(1*(1+x)*(1+x+x^2)*(1+x+x^2+x^3)*(1+x+x^2+x^3+x^4)))

we get the output
$$x^{10} + 4 \, x^{9} + 9 \, x^{8} + 15 \, x^{7} + 20 \, x^{6} + 22 \, x^{5} + 20 \, x^{4} + 15 \, x^{3} + 9 \, x^{2} + 4 \, x + 1$$
which gives 9 as your answer.
Edit: The full list of permutations with $2$ inversions. I have highlighted all occurrences of a larger number occurring to the left of a smaller. The ones accounted for in the question are checked $\checkmark$.
$$\begin{array}{c}\text{Permutations}\\
\begin{array}{|ccccc|c|}\hline
\boxed{2} & \boxed{3} & 1 & 4 & 5 &\checkmark\\
\boxed{2} & 1 & \boxed{4} & 3 & 5&\\
\boxed{2} & 1 & 3 & \boxed{5} & 4&\\
1 & \boxed{3} & \boxed{4} & 2 & 5&\checkmark\\
1 & \boxed{3} & 2 & \boxed{5} & 4&\\
1 & \boxed{4} & 2 & 3 & 5&\checkmark\\
1 & 2 & \boxed{4} & \boxed{5} & 3&\checkmark\\
1 & 2 & \boxed{5} & \boxed{3} & 4&\checkmark\\
\boxed{3} & 1 & 2 & 4 & 5&\checkmark\\\hline
\end{array}\end{array}$$
