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Thread: Alternating arrangements at a table

  1. #1
    Member jacs's Avatar
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    Alternating arrangements at a table

    Hello, and thanks in advance.

    Four adults and 4 children are to be seated around a circular table. A particular child cannot sit next to any adult and a particular adult cannot sit next to any child.
    Find how many such arrangements are possible.


    not sure i have done this correctly and was hoping for some guidance.

    Let the child that cannot sit next to adult be C1 and the adult that cannot sit next to any child be A1

    Seat C1 in position 1 leaving only positions 4, 5, 6 for A1 = 3!

    Seat A1
    this leaves 3! positions for the remaining children
    and 2! for the remaining adults
    Total = 3! x 3! x 2!
    = 72 ways

    would appreciate confirmation if this is correct number of ways (i have no solution set)
    or if i am completely on the wrong track
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  2. #2
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    Re: Alternating arrangements at a table

    I get 72 as well.

    you'll have an arrangement

    $A A_1 A (A C) C C_1 C (A C)$

    There are 3 ways to choose the first A

    2 ways to choose the 2nd A

    3 ways to choose the first C

    2 ways to choose the second C

    Then you have the choice of either A or C for the 3rd slot. Having chosen this the 7th slot is fixed.

    3 x 2 x 3 x 2 x 2 = 9 x 8 = 72
    Thanks from jacs
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