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Probablity is greater than 6/{(n-2)^3}

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September 5, 2010
probabilityarithmetic sequencecombinatorics unsolvedcombinatorics

Problem Statement

Five distinct numbers are drawn successively and at random from the set {1,⋯ ,n}\{1, \cdots , n\}. Show that the probability of a draw in which the first three numbers as well as all five numbers can be arranged to form an arithmetic progression is greater than 6(n−2)3\frac{6}{(n-2)^3}
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