In a permutation a1.....an of n distinct integers, an inversion is a pair (ai,…

2003

In a permutation a1.....an of n distinct integers, an inversion is a pair (ai, aj) such that i < j and ai > aj. If all permutations are equally likely, what is the expected number of inversions in a randomly chosen permutation of 1.....n ?

  1. A.

    n(n - 1)/2

  2. B.

    n(n - 1)/4

  3. C.

    n(n + 1)/4

  4. D.

    2n[log2 n]

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Correct answer: B

Solution (indicator variable method):

  • For each pair of positions i and j with i < j, define an indicator variable X_{ij} that equals 1 if the pair forms an inversion (the element in position i is greater than the element in position j) and 0 otherwise.

  • Because all permutations are equally likely, for any fixed pair i < j the two relative orders are equally likely, so P(X_{ij} = 1) = 1/2 and therefore E[X_{ij}] = 1/2.

  • There are exactly n(n - 1)/2 distinct pairs i < j. The total number of inversions is the sum of the X_{ij}, so by linearity of expectation the expected total is

E[number of inversions] = (number of pairs) × E[X_{ij}] = [n(n - 1)/2] × 1/2 = n(n - 1)/4.

Quick check: for n = 3 there are 6 permutations with inversion counts 0,1,1,2,2,3; the average is (0+1+1+2+2+3)/6 = 9/6 = 3/2, which matches n(n - 1)/4 = 3·2/4 = 1.5.

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