Saturday, December 02, 2006

Not a joke. But very funny anyway.

Nevermind.

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Puzzle of the day from perplexus.info:
Alternating Differences (Difficulty: 3 of 5)

Let X1, X2, X3, ..., Xn be a permutation of the integers 1,2,3,...,n.

Consider the sum:
abs(X1-X3) + abs(X2-X4) + abs(X3-X5) + ... + abs(Xn-2-Xn).

What is the mean value of this sum taken over all possible permutations?

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