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Thread: Six people permutation

  1. #1
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    Six people permutation

    We have 6 people, 3 men, and 3 women.

    On how many ways, can we sat them at a round table, on 6 chairs, but they must sit in this order: man, woman,man,woman... So the neighbour must be the opposite sex.

    Is this correct?

    $\displaystyle \frac{3!*3!}{6}$

    the six in the denominator is because it's a round table, and we can rotate them for one place, and would be the same so it's 6 for six places.
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  2. #2
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    Re: Six people permutation

    Quote Originally Posted by Nforce View Post
    We have 6 people, 3 men, and 3 women.
    On how many ways, can we sat them at a round table, on 6 chairs, but they must sit in this order: man, woman,man,woman... So the neighbour must be the opposite sex.
    Is this correct?
    $\displaystyle \frac{3!*3!}{6}$
    No it is not correct. The correct answer is $\displaystyle 2!\cdot 3!$
    You tell us why.
    Thanks from Nforce
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  3. #3
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    Re: Six people permutation

    Ok, I made this on a long way, first man has 6 choices, second 2 and third one, and for the woman 3!, and because it's round we devided by 6.
    and, get...

    $\displaystyle \frac{6*2*1*3!}{6} = 12$

    it's tricky, but how did you get that 2! * 3! so fast?
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  4. #4
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    Re: Six people permutation

    Quote Originally Posted by Nforce View Post
    Ok, I made this on a long way, first man has 6 choices, second 2 and third one, and for the woman 3!, and because it's round we devided by 6.
    and, get...
    $\displaystyle \frac{6*2*1*3!}{6} = 12$
    it's tricky, but how did you get that 2! * 3! so fast?
    The way you did it works. It is just different.
    To seat the six at the table in any order is $\displaystyle \frac{6!}{6}=5!$.
    There are $\displaystyle \frac{3!}{3}=2!$ ways to seat the men in every other chair.
    There are $\displaystyle 3!$ ways to seat the women at the remaining, now ordered, table.
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