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1 | tourist | 3690 |
2 | jiangly | 3647 |
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5 | Geothermal | 3569 |
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7 | Radewoosh | 3509 |
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1 | maomao90 | 174 |
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7 | -is-this-fft- | 152 |
8 | Petr | 146 |
8 | orz | 146 |
10 | BledDest | 145 |
Название |
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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 $$$N! (N-1) \frac{2}{N(N-1)} \sum_{i=1}^{N} \sum_{j=1+1}^{N} (j-i) = N! \frac{2}{N} \sum_{i=1}^{N} \sum_{j=1+1}^{N} (j-i)$$$