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Cominatorial Probability

  1. Oct 22, 2007 #1
    1. The problem statement, all variables and given/known data

    If Sam and Peter are among n men who are arranged at random in a line, what is the probability that exactly k men stand between them?"

    2. Relevant equations

    Not sure.

    3. The attempt at a solution

    This is driving me insane. I listed out the possibilities for n = 3, and it looked to me like Peter and Sam could surround n men


    ways. Since there are n! possible lines, I said:

    p = 2(n-k-1)/n!

    But the book says

    p = 2(n-k-1)/n(n-1)

    The frustrating thing is that our answers are equal for n=3, but obviously not for higher values of n. I started trying to do it concretely for n=4, but there were so many possibilities that I got too confused.

    It's just killing me, because I know the formula I came up with for n=3 was just a guess based on my empirical observation of the outcomes. I don't know how to do this problem "theoretically."

    What am I doing wrong?

  2. jcsd
  3. Oct 22, 2007 #2


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    Homework Helper

    n! is the number of permutations, but you need to look at the combinations.
  4. Oct 22, 2007 #3
    Well, in a line doesn't the ordering matter? So wouldn't it be permutations?

    I mean, if the total number of possible lines were n choose n, their would only be 1 possible line!!!
  5. Oct 22, 2007 #4


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    Homework Helper

    Yes, it's permutations you want. But your 2n(n-k-1) is the number of ways Sam and Peter can stand. There are (n-2) other men that you have to put in line with Sam and Peter before you can divide by n!.
  6. Oct 22, 2007 #5
    Thank you!!!!!!! I finally get it.

    Wow, combinatorics is a brain trip. I've taken high level differential equations, geometry, and all kinds of signals stuff in EE, but these counting problems crack my skull like nothing else! They are so deceptive.
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