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Probablity of fixed points in permutations

  1. Feb 14, 2008 #1
    Randomly permute (1,...,n). What is the probability that exactly i points are fixed?

    I think it should be

    [tex]\binom{n}{i}\frac{(n-i-1)!}{n!}[/tex]

    Is it right?

    If so, is the expected number of fixed points (I know it's 1):

    [tex]\sum_{i=0}^{n}i\binom{n}{i}\frac{(n-i-1)!}{n!}[/tex]

    But it doesn't sum to 1, I think
     
    Last edited: Feb 14, 2008
  2. jcsd
  3. Feb 14, 2008 #2

    EnumaElish

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    How is (n-i-1)! defined when i = n?
     
  4. Feb 14, 2008 #3
    It isn't. Nevermind. I was approaching the problem in a unnecessarily hard way.
     
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