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Probability problem (#of ways to seat a couple) *edit: solved*

  1. Mar 25, 2013 #1
    update:
    *edit: solved! I counted everything twice.


    1. The problem statement, all variables and given/known data

    How many ways are there to seat 10 people, consisting of 5 couples, in a row
    of seats (10 seats wide) if all couples are to get adjacent seats?





    3. The attempt at a solution
    I'm trying to look calculate it at a couple level.

    10 seats available, 5 couples
    _ _ _ _ _ _ _ _ _ _

    first couple has 9 choices to choose from They can't choose seat # 10 or else they will not sit next to each other. they can also swap seats, so that will be 9 x 2 = 18 choices.

    second couple will have 7 seats to choose from. They can trade seats, so that will be 7 x 2 = 14

    third couple has 5 seats, trade seats an yield 10.

    fourth couple has 3 seats, = 6

    5th couple has 2 seats left, they can trade once, so that is 2 possibilities

    I did 18 x 14 x 10 x 6 x 2 = 30,240. answer key says 3840 though?
     
    Last edited: Mar 25, 2013
  2. jcsd
  3. Mar 25, 2013 #2

    CompuChip

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    Science Advisor
    Homework Helper

    Since you have already found the correct answer, I will share another way to get there.
    As each couple has to sit next to each other, you basically have 5 two-seat blocks to distribute the 5 couples over. The number of ways in which this can be done is just 5! because every permutation of the couples will give a different assignment of couples to two-seat blocks. Then within each block you can swap the two persons, so you get [itex]5! \cdot 2^5[/itex] possible ways.
     
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