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  1. #1
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    round table

    4 married couples sit at a round table. How many ways can they sit if each husband must sit opposite his wife?


    3!3.2! is my answer.
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    Re: round table

    Quote Originally Posted by kastamonu View Post
    4 married couples sit at a round table. How many ways can they sit if each husband must sit opposite his wife?
    3!3.2! is my answer.

    Seat couple A at the table.
    Now there are six ways to seat couple B.
    Then there are four ways to seat couple C.
    Finally there are two ways to seat couple D.
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    Re: round table

    3!.16 . If we think four couples as one person we can place them in 3! ways. 2 couples will be ordered between themselves and there are 4 couples this makes 16. But your nswer is better. Many Thanks.
    Last edited by kastamonu; February 18th 2013 at 07:02 AM.
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    Re: round table

    Quote Originally Posted by kastamonu View Post
    3!.16 . If we think four couples as one person we can place them in 3! ways. 2 couples will be ordered between themselves and there are 4 couples this makes 16.

    I don't know where you are getting your model, but it is wrong.
    Draw a regular octagon. There vertices are the eight chairs.
    It is not ordered until one couple is seated.
    Seat couple A anywhere, but man and wife are at opposite vertices.
    Now the table is ordered.

    There are six seats left. Seat wife B in any one of them. Then man B has only one place to be seated, at the opposite vertex from wife B.

    6\cdot 4\cdot 2=48.
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    Re: round table

    What happens if we want all the couples to sit seperately?
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    Re: round table

    There are 4 married couples. If we think them as 1 person they can be arranged in 3! They will change places between themselves in 2! ways. TRhere are 4 couples and this makes 2!2!2!2!3!
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    Re: round table

    Quote Originally Posted by kastamonu View Post
    What happens if we want all the couples to sit seperately?

    Always, always, start a new thread for a new question.

    Using inclusion/exclusion \sum\limits_{k = 0}^4 {\left( { - 1} \right)^k \binom{ 4}{k}  \cdot 2^k \left( {7 - k} \right)!}
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