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Math Help - Permutations of [n]

  1. #1
    Member oldguynewstudent's Avatar
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    Permutations of [n]

    How many permutations of [n] are possible in which no even numbers and no odd numbers are adjacent?

    Take [5] = 1 2 3 4 5. The ways to arrange the even numbers are 2 4 and 4 2 as long as stay in the even number slots. Then the way to arrange the odd numbers are 1 3 5; 1 5 3; 3 1 5; 3 5 1; 5 1 3; and 5 3 1. The number of permutations of [5] in which no even numbers and no odd numbers are adjacent is 2! * 3! or 2P2 times 3P3.

    So I see the formula as two cases if n is even then it would be

    _{\frac{n}{2}}P_{\frac{n}{2}} * _{\frac{n}{2}}P_{\frac{n}{2}}

    If n is an odd number then

    _{\frac{n+1}{2}}P_{\frac{n+1}{2}} * _{\frac{n-1}{2}}P_{\frac{n-1}{2}}

    This seems too complicated.

    Is there a way to calculate the total number of permutations n! and subtract the cases where the even numbers would be adjacent and the cases where the odd numbers would be adjacent?
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  2. #2
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    Those are almost correct.

    In the case n being even we need to double your formula.
    Consider [4], 1432 is a correct string. But so is 4123.
    You did not take that into consideration.

    You are correct in the case n is odd.
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  3. #3
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Plato View Post
    Those are almost correct.

    In the case n being even we need to double your formula.
    Consider [4], 1432 is a correct string. But so is 4123.
    You did not take that into consideration.

    You are correct in the case n is odd.
    Thank you again. You're explanations are well appreciated!
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