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A simple permutation question or not so simple

  1. Jun 14, 2005 #1
    How many ways can 2 men and 3 women be seated in a row such that no 2 men are sitting beside each other?

    Now I have always had a problem with overthinking these kinds of questions. I'll usually write something down but then doubt myself.

    What I did was simply did 2! * 3!.
    3 * 2 * 2 * 1 * 1
    W M W M W
    That's by alternating women with men.

    But here comes me overthinking the problem again.
    I could altenate with men first, and then have 2 women sitting beside each other.
    M W M W W right?
    Or W W M W M....

    Ahhh I feel like an idiot.
  2. jcsd
  3. Jun 14, 2005 #2


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    In each of these arrangements, there are 2!3! ways to choose the men and women, so there are 72 ways to arrange them. (Unless, of course, I've made a mistake somewhere.)
  4. Jun 14, 2005 #3
    What if I were to up the ante of men and women to 4 and 5, respectively.
    During an exam, I shouldn't be writing all the possible permutations for the problem.

    I would get 4!5! for each arrangement, but how do I know how many arrangements there is going to be?
  5. Jun 14, 2005 #4
    you must think of the 2 men as 1 with 2 permutation for MaMb and MbMa.
  6. Jun 15, 2005 #5
    I should of mentioed that each person is distinguishable from another. There are no two that are a like.
  7. Jun 15, 2005 #6
    What I got was,

    There are 9 in total. Each arrangement is 5!4!.
    Therefore 5!*4!*9 is the number of ways this can be done.

    Is this coorect?
  8. Jun 15, 2005 #7
    here is a solution that I found, but uses a round table instead of a row.

    How many ways can 5 man and 7 women be seated at a round table with
    no 2 men next to each other?
    Solution. First place the women in 6!. Now there are 7C5 ways to pick
    5 spots for the men so that they are not adjancent. Finally, in each
    of these 5 spots, the men can be placed in 5! ways. Hence, there are
    5!6!7C5 = 1814400.

    What I have is 5!4!5C4
  9. Jun 15, 2005 #8
    Do it this way:
    First place the women in order. This is 5!. Then place the men in the spaces between the women. There are 6 spaces (including the ends) so you have C(6, 4) * 4! ways to place the men, and 5! * C(6, 4) * 4! altogether.

    Now I'll check by listing:

    110000 wwmwmwmwm
    101000 wmwwmwmwm
    100100 wmwmwwmwm
    100010 wmwmwmwwm
    100001 wmwmwmwmw
    011000 mwwwmwmwm
    010100 mwwmwwmwm
    010010 mwwmwmwwm
    010001 mwwmwmwmw
    001100 mwmwwwmwm
    001010 mwmwwmwwm
    001001 mwmwwmwmw
    000110 mwmwmwwwm
    000101 mwmwmwwmw
    000011 mwmwmwmww

    So there are indeed C(6, 2) = 15 ways to arrange the men and women before permuting.
  10. Jun 16, 2005 #9

    awesome man, this is exactly how my prof wanted us to do it. thanks
  11. Jun 18, 2005 #10
    Don't forget you can always have two of the women sit on the men's lap. :!!)
    Last edited: Jun 18, 2005
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