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Math Help - number of derangements

  1. #1
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    number of derangements

    How is it possible that D8 > 8! ?
    8! is the number of all possible permutations of 8 elements, and D8 is the number of permutations such that no element is in its position.
    So {set of derangements} \subset {set of permutations}.
    How can there be more derangements than permutations of the same number of elements?
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  2. #2
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    Re: number of derangements

    It isn't possible. What values do you get for D_8 and 8! ?

    According to Wikipedia, Derangement - Wikipedia, the free encyclopedia, 8!= 40320 while D_8= 14833.
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  3. #3
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    Re: number of derangements

    Well this is embarrassing...
    I had somehow read the value on my calculator as 4320.

    Thank you.
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  4. #4
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    Re: number of derangements

    Quote Originally Posted by Yuuki View Post
    How is it possible that D8 > 8! ?
    8! is the number of all possible permutations of 8 elements, and D8 is the number of permutations such that no element is in its position.
    So {set of derangements} \subset {set of permutations}.
    How can there be more derangements than permutations of the same number of elements?
    Here is a bit more. Look at the webpage in reply #2.

    It is common to use the approximation, \mathcal{D}(N)\approx \frac{N!}{e}~.

    That shows the impossibility.
    Thanks from Yuuki
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