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Committee combination math problem

  1. Dec 18, 2003 #1
    For a commitee, 3 people out of 4 lawyers, 1 minister and 3 retailers are to be chosen. 1 person in the commitee must be a retailer, how many ways are there to choose the commitee?

    1st: There must be a retailer thus we have 3 choices
    2nd: Out of the remaining 7 people, the possible combinations are C(7,2)

    So there must be 3C(7,2) possibilities. Is it right?
  2. jcsd
  3. Dec 18, 2003 #2


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    How many times did you count the cases with two retailers?

    This problem is probably easier to solve by first solving its complement: how many committees don't have any retailers?
  4. Dec 18, 2003 #3
    Then you would have 5 people remaining. I would have C(5,3). That is if there were no retailers. With retailers, I would have, C(8,3). Thus C(8,3)-C(5,3) equals 46...OMG I WISH I ONLY KNEW THAT WHEN I WAS TAKING THE TEST!!! I NEED MORE PRACTICE, thanks Hurkyl!:smile:
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