Nice drawing :D

What's that character in brackets after summations before i (?-i)

I don't know if this convince you.

Let's assume the difference is taking an absolute sign. There are $$$n!$$$ permutation of len $$$n$$$. For one permutation there are $$$n-1$$$ adjacent difference. The probability of getting one pair $$$(x,y)$$$ or $$$(y,x)$$$ , having the difference $$$|x-y|$$$ is $$$\frac{2}{n(n-1)}$$$

So the sum will be $$$N! (N-1) \frac{2}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1+1}^{N} (j-i) = N! \frac{2}{N} \sum_{i=1}^{N} \sum_{j=1+1}^{N} (j-i)$$$