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Math Help - Hard combinatorics problem

  1. #1
    MHF Contributor alexmahone's Avatar
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    Hard combinatorics problem

    In how many ways can the elements of [n] be permuted if 1 is to precede 2 and 3 is to precede 4?

    [n] is the set of integers from 1 to n.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: Hard combinatorics problem

    Thinking...

    The problem is meaningful only for n >= 4.
    For n = 4, there are 6 possible permutations: (1, 2, 3, 4), (1, 3, 2, 4), (1, 3, 4, 2), (3, 4, 1, 2), (3, 1, 4, 2) and (3, 1, 2, 4).
    For n = 5, '5' can go into any of the five 'gaps'. There are 5*6 = 30 possible permutations.

    The numbers greater than 4 will have to placed into some or all of the five 'gaps'.
    Last edited by alexmahone; August 17th 2011 at 04:12 PM.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Hard combinatorics problem

    (Not entirely my own.)

    The problem is meaningful only for n >= 4.
    When n = 4, there are 6 possible permutations: (1, 2, 3, 4), (1, 3, 2, 4), (1, 3, 4, 2), (3, 4, 1, 2), (3, 1, 4, 2) and (3, 1, 2, 4).
    When n = 5, '5' can go into any of the five 'gaps' of n = 4. There are 5*6 = 30 possible permutations.
    When n = 6, '6' can go into any of the six 'gaps' of n = 5. There are 6*30 = 180 possible permutations.
    ... and so on.
    So, for n > 4, there are n(n-1)...5 * 6 possible permutations. We rewrite 6 as 4!/4.
    So, the number of possible permutations is n(n-1)...5*4!/4 = n!/4.
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