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The set of 2n numbers {1,2,3,...,2n} are split randomly into two subsets of n elements. Then we sort each subset. Denote the two ordered subsets by $A={a_1 ,a_2 ,...,a_n } (a_1 <a_2 <...<a_n )$ and $B={b_1 ,b_2 ,...,b_n } (b_1 >b_2 >...>b_n )$. What is the expectation E(S) of S, where $S=\sum_{i=1}^n(|a_i-b_i|$?

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If we pick $x,y$ and $j$

what is the probability that $a_i=x,b_i=y$?

without loss of generality we have $x<y$.

Then it is $\dfrac{\binom{x-1}{i-1}\binom{y-x-1}{2i-x}}{2^y}$

So our result is $2\sum\limits_{i=1}^n \sum\limits_{x<y}\dfrac{(y-x)\binom{x-1}{i-1}\binom{y-x-1}{2i-x}}{2^y}$

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  • $\begingroup$ So you can calculate in cubic complexity, $\endgroup$ – Jorge Fernández Hidalgo Jan 1 '17 at 2:16
  • $\begingroup$ Can you explain a little bit more details please? I cannot follow every step. Thanks. $\endgroup$ – Della Jan 16 '17 at 13:23

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