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A question about permutations

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data

    The coach from a soccer team of 15 players must select 11 players for the start of a game.

    ) Before the game all players line up in a straight line for a team photograph. If 2 players,
    Michaela and Aleah must be together, then how many different arrangements can be made
    for the picture?

    2. Relevant equations
    N/A


    3. The attempt at a solution

    2*13!


    The correct answer is 2*14! but I don't understand why they chose 14!, since there are only 13 players left after you isolate Michaela and Aleah, who can either be MA or AM.

    Nevermind I figured it out.
     
    Last edited: Aug 4, 2013
  2. jcsd
  3. Aug 4, 2013 #2

    CAF123

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    Gold Member

    Just in case anyone else is interested, the 2*13! only takes into account one possibility for the 'block' of the two girls (either MA or AM). There are 15 people, so this block may be put in 14different positions i.e this increases the number of possible different photographs to be 2*13!*14 = 2*14!.
     
  4. Aug 5, 2013 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You treat "Michaela-Aleah" as a single person. That leaves 14 "persons" to place, the 13 other people and the "Michaela-Aleah" pair: 14!. And, of course there are the two different ways of ordering that pair, "Michaela-Aleah" or "Aleah-Michaela".

    Another way to think about it: Withdraw Aleah from the group. There are now 14 people and so 14! ways to order them. We now can decicde to put Aleah to the left or right of Michaela: 2 ways to do that so (14!)(2).
     
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