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Homework Help: Permutations - arrangement of a boat crew of 8 women

  1. Apr 5, 2013 #1
    1. The problem statement, all variables and given/known data
    How many ways can a boat crew of 8 women be arranged if 3 of the women can only row on the bow side and 2 can only row on the stroke side?

    2. Relevant equations

    3. The attempt at a solution
    I simply did 8!/5! (5!/3!)3! which is simply 3!(8!/3!)

    The correct answer is 1728.
    Last edited: Apr 5, 2013
  2. jcsd
  3. Apr 5, 2013 #2


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    Homework Helper

    And 3!(8!/3!) is simply 8!, which is the total number of ways you can arrange the crew without taking the constraints into account.

    Let's try to get some intuition first... let B denote someone who can only row on the bow side, S on the stroke side, and A someone who can be on either side.
    One possible arrangement is:

    A A
    B A
    B S
    B S

    You need to realise there are two possible ways of permuting this
    (1) you can move the A's around on the left hand side or the right hand side, e.g.
    B A
    A S
    B A
    B S

    So you will have to figure out the number of permutations this gives you.

    (2) in addition, all the rowers are individuals whose order matters. So for any variation you get from the above, you can move the B's, the S's and the A's, e.g. if we number the ladies then the first layout above can either correspond to
    A1 A2
    B1 A3
    B2 S1
    B3 S2

    or to

    A2 A3
    B2 A1
    B3 S2
    B1 S1

    or any other combination which again leads to (A, B, B, B; A, A, S, S) when you remove the numbers.
  4. Apr 5, 2013 #3


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    You can try this also:

    The four seats on the bow side will "select three seats" in which the the three women will sit. The order does not count. Similar for the stroke side, which will "select two seats", and the order does not count. After women are seated on each of the sides, there are three women to be seated in 3 places in any order.
  5. Apr 6, 2013 #4
    You can also reduce the full permutation successively by eliminating the proportion of "wrong" permutations.

    So first considering the bow-siders, the first could have been in 8 seats of which 4 are correct (multiply by 4/8); the second could have been in 7 (remaining) seats of which 3 are correct, then 2 from 6 seats, and then on the stroke-siders, 4 from 5 seats, and 3 from 4 seats.
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