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  1. #1
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    in how many ways can...

    In how many ways can you seat 4 couples at a round table if:
    1) each couple sits together?
    2) no couples sits together?

    I got 12 for the first...I highly doubt it's right though. Any tups? I don't want you to actually solve it all out for me, but pointing me in the right direction would be nice
    thanks!
    I know that: # of circular combinations = # of linear combinations / # of circle combinations that can occur from each linear combination.
    I just can't really figure out exactly how to apply that to these problems...
    =/
    Last edited by Sconts; May 2nd 2007 at 04:38 PM. Reason: correcting a typo that severely altered the message
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  2. #2
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    In how many ways can you seat 4 couples at a round table if:
    1) each couple sits together?
    2) no couples sits together?

    you have 8 people. 8 chairs

    1. If the couple has to sit together then...the first chair has 8 possible ways, but the next one can only have 1 possible way which is the persons significant other. You have 6 chairs left and the 3rd chair now has 6 possible ways, and the next one to it has only 1...and etc... so...

    8 * 1 * 6 * 1 * 4 * 1 * 2 * 1 = 384 ways.

    2. the total possible ways to seat these 4 couples with no restrictions...
    8! = 40320

    if no couples sits together....then you take the total possible ways minus the possible ways of them sitting together.

    8! - 384 = 39936
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  3. #3
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    Hello, Sconts!

    Here's part (1) . . .


    In how many ways can you seat 4 couples at a round table if:
    (1) each couple sits together?
    (2) no couples sits together?

    1) If the couples must be adjacent, duct-tape the pairs together.
    Then we have four "people" to arrange around a round table.

    The first couple can sit anywhere
    There are 3! = 6 ways to seat the other three couples.

    But each couple can be seated in two ways: Aa or aA.
    Hence, there are: 2^4 = 16 possible orderings.

    Therefore, there are: .6 16 .= .96 possible seatings.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Using smoothi's approach . . .

    The first person can sit anywhere; call that person/position #1.
    . . Let us move clockwise around the table.
    Position #2 must be occupied by 1's spouse: 1 choice.

    Position #3 can be occupied by any of the six remaining people: 6 choices.
    Position #4 must be occupied by 3's spouse: 1 choice.

    Position #5 can be occupied by any of the four remaining people: 4 choices.
    Position #6 must be occupied by 5's spouse: 1 choice.

    Position #7 can be occupied by either of the two remaining people: 2 choices.
    Position #8 must be occupied by 7's spouse: 1 choice.

    Hence, there are: .1 x 6 x 1 x 4 x 1 x 2 x 1 .= .48 ways.


    If we move counterclockwise around the table, we find 48 more arrangements.
    . . These are mirror-images of the first 48 arrangements.


    Therefore, there are: .48 + 48 .= .96 possible seatings.

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