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Math Help - Another hard combinatorics problem

  1. #1
    MHF Contributor alexmahone's Avatar
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    Another hard combinatorics problem

    In how many ways can the elements of [n] be permuted if 1 is to precede both 2 and 3?

    My solution: {1, 2, 3} can be permuted in 6 ways. Out of them, only 2: (1, 2, 3) and (1, 3, 2) are acceptable.
    By symmetry, all the 6 permutations must occur equal number of times. Total number of permutations is n! So, the acceptable number of permutations is n!/3.

    Is this correct?
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  2. #2
    Member Traveller's Avatar
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    Re: Another hard combinatorics problem

    Yes.
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