How can I show that the number of inversions of a permutaion of $\displaystyle \{1, 2, \dots , n\} $ is $\displaystyle n(n-1)/2 $
I guess by inversion you mean transpositions i.e. $\displaystyle 2$-cycles. Note that any inversion has form $\displaystyle (ij)$ where $\displaystyle i<j$ and furthermore this is unique. Therefore the number of such inversions is $\displaystyle {n\choose 2} = \frac{n(n-1)}{2}$.