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Permutations and arranging order

  1. Jul 10, 2007 #1
    Hi, I was answering what I thought was an easy problem and I got it wrong, but not sure why. Please give me an insight.

    Problem: 12 juniors are ordered (in a line) for a drill. What's the probability (assuming all arrangments are random) of Dave standing next to Beth?

    My reasoning: There are 12! possible arrangments of juniors. The number of arrangments where Dave is next to Beth is 2 (number of ways Beth can stand next to Dave, DB - BD) multiply by all possible arrangments for the remaning juniors, i.e. 10!. So, the probability is then 2x10! / 12!.

    Well, that's wrong - 2 must be multiplied by 11!, not 10!, but I don't understand why - there are 10 juniors left to be ordered in any way to be combined with Beth and Dave. I realize that if I write it out on a paper, I'll count 11! of them. But I don't want to simply memorize the formula for this type of problem, I want understand the reasoning. Why 11! ?


  2. jcsd
  3. Jul 10, 2007 #2


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    You have to tell Beth where to stand before you place David next to her.
  4. Jul 10, 2007 #3


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    There are 10! ways for the others to line up, and thus 10! ways for Dave to be first, Beth second, and everyone else in some order behind them. If Dave and Beth can be in any order that is 2*10! ways. But Dave and Beth don't have to be together at the front of the line -- they could be anywhere. There are 2*10! ways there could be one person in front of the two (and 9 behind), 2*10! that there are two people in front of them, etc., for a total of 11 * (2 * 10!) ways, or 2 * 11! ways.

    Edit: or what EnumaElish said.
  5. Jul 10, 2007 #4


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    OK, so you have 2 ways for Beth to stand next to Dave, BD and DB, and thus have 10! ways of arranging the remaining children. However, there are 11 ways of arranging the combination of Beth and Dave amongst the other children. So, there are 2*11*10! ways of arranging the children so that B and D are next to one another.
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