Saiyan_1991's blog

How to prove it?
By Saiyan_1991, 4 years ago, In English

If you can't prove it please give an upvote so that higher ranked user can see this post.

And as this is a alt account I don't care about contribution. I just need help. Thank you.

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4 years ago, # |
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Nice drawing :D

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4 years ago, # |
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What's that character in brackets after summations before i (?-i)

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4 years ago, # |
  Vote: I like it +19 Vote: I do not like it

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 $$$N! (N-1) \frac{2}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1+1}^{N} (j-i) = N! \frac{2}{N} \sum_{i=1}^{N} \sum_{j=1+1}^{N} (j-i)$$$

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