Inversions of a permutation

**hcir614** · September 28th, 2008, 12:18 PM

How can I show that the number of inversions of a permutaion of $\{1, 2, \dots , n\}$ is $n(n-1)/2$

**ThePerfectHacker** · September 29th, 2008, 01:22 PM

Originally Posted by hcir614

How can I show that the number of inversions of a permutaion of $\{1, 2, \dots , n\}$ is $n(n-1)/2$

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

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