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Math Help - Number of permutations

  1. #1
    MHF Contributor alexmahone's Avatar
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    Number of permutations

    In how many ways can the elements of [n] be permuted so that the sum of every two consecutive elements in the permutation is odd?

    [n] is the set of integers from 1 to n.

    My attempt:

    If n is even:
    n\frac{n}{2}\left(\frac{n}{2} - 1\right)\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\left(\frac{n}{2}-2\right)...=2\left(\frac{n}{2}!\right)^2

    If n is odd, we must start with an odd element:
    \left(\frac{n+1}{2}\right)\left(\frac{n-1}{2}\right)\left(\frac{n-1}{2}\right)\left(\frac{n-3}{2}\right)\left(\frac{n-3}{2}\right)... =\left(\frac{n+1}{2}\right)\left(\frac{n-1}{2}!\right)^2

    Is this right?
    Last edited by alexmahone; August 18th 2011 at 12:57 PM.
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  2. #2
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    Re: Number of permutations

    Quote Originally Posted by alexmahone View Post
    In how many ways can the elements of [n] be permuted so that the sum of every two consecutive elements in the permutation is odd?
    [n] is the set of integers from 1 to n.

    My attempt:
    If n is even:
    n\frac{n}{2}\left(\frac{n}{2} - 1\right)\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)\left(\frac{n}{2}-2\right)...=2\left(\frac{n}{2}!\right)^2

    If n is odd, we must start with an odd element:
    I agree with you as far as that goes.
    But in is odd it should be: because start with an odd and alternate
    \left[ {\left( {\frac{{n + 1}}{2}} \right)!} \right] \cdot \left[ {\left( {\frac{{n - 1}}{2}} \right)!} \right]
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Number of permutations

    Quote Originally Posted by Plato View Post
    I agree with you as far as that goes.
    But in is odd it should be: because start with an odd and alternate
    \left[ {\left( {\frac{{n + 1}}{2}} \right)!} \right] \cdot \left[ {\left( {\frac{{n - 1}}{2}} \right)!} \right]
    Actually, that's exactly what I got. But there was a problem with the latex (which I've now fixed).

    \left(\frac{n+1}{2}\right)\left(\frac{n-1}{2}!\right)^2 = \left(\frac{n+1}{2}!\right)\left(\frac{n-1}{2}!\right)
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