olymmath

RU-ALLRUSSIAN 1976 Problem 12

    n is a positive integer. A universal sequence of length m is a sequence of m integers each between 1 and n such that one can obtain any permutation of 1, 2, ..., n by deleting suitable members of the sequence. For example, 1, 2, 3, 1, 2, 1, 3 is a universal sequence of length 7 for n=3n = 3 . But 1, 2, 3, 2, 1, 3, 1 is not universal, because one cannot obtain the permutation 3, 1, 2. Show that one can always obtain a universal sequence for n of length n2−n+1n^2 - n + 1 . Show that a universal sequence for n must have length at least n(n+1)/2n(n + 1) / 2 . Show that the shortest sequence for n=4n = 4 has 12 members. [You are told, but do not have to prove, that there is a universal sequence for n of length n2−2n+4n^2 - 2n + 4 .]

    Dataset text NuminaMath olympiads. Quoted with source.

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