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Math Help - Circular table permutations problem

  1. #1
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    Circular table permutations problem

    Hey, i have problem understanding how to get result for circular permutations where i have repeating elements.
    I have 10 people in circle, 3 man and 7 women. How can i get number of permutations?

    Result in book is 36.
    I tried
    (9!/2!+7!)+(9!/3!+6!)=120 which is wrong.
    Last edited by PJani; May 11th 2013 at 07:02 AM.
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  2. #2
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    Re: Circular table permutations problem

    Quote Originally Posted by PJani View Post
    Hey, i have problem understanding how to get result for circular permutations where i have repeating elements. I have 10 people in circle, 3 man and 7 women. How can i get number of permutations?

    Result in book is 36.
    That problem as you have posted it is absolutely incomplete.

    As stated the answer is simply 9!.

    There must be many more conditions placed on the groups of men and women.

    Please add in those further conditions. That is the only way the answer 36 makes any sense.
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  3. #3
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    Re: Circular table permutations problem

    Heh sory, my mistake!
    How can i get number of permutations where i distinguish persons only by gender?
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  4. #4
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    Re: Circular table permutations problem

    Quote Originally Posted by PJani View Post
    Heh sory, my mistake!
    How can i get number of permutations where i distinguish persons only by gender?
    In the future, please try to be clear and complete.

    Place a female anywhere at the table.
    Now you have a string ffmmmmmmm to arrange: \frac{9!}{2!\cdot 7!}=? ways.

    Start at the seated female's right and seat each of those strings counter-clockwise.
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  5. #5
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    Re: Circular table permutations problem

    Why do you start with females? And why is \frac{9!}{2!\cdot 7!}=? and not \frac{9!}{3!\cdot 6!}=?
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