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Math Help - Random order of numbers from 1 to 8 - at least six are in their correct slot.

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    Random order of numbers from 1 to 8 - at least six are in their correct slot.

    Determine the probability of that if you randomly order the numbers from 1 to 8 that at least 6 of them are in their correct slot.

    This has to do with probability and factorials. Steps shown would be helpful.
    Last edited by mr fantastic; May 2nd 2010 at 01:40 PM. Reason: Re-titled.
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    Quote Originally Posted by sapphire View Post
    Determine the probability of that if you randomly order the numbers from 1 to 8 that at least 6 of them are in their correct slot.
    This has to do with probability and factorials. Steps shown would be helpful.
    If one rearranges a string of n different items so that no one item appears in its original position that is called a derangement.
    The number of derangements of n items is given by  D(n) = n!\left( {\sum\limits_{K = 0}^n {\frac{{\left( { - 1} \right)^k }}{{k!}}} } \right) \approx \frac{{n!}}{e}.

    So to rearrange 12345678 so that at least six numbers remain in the correct position can be done
    in \sum\limits_{k = 6}^8 {\binom{8}{k}} D(8-k) number of ways.
    Last edited by Plato; May 2nd 2010 at 02:38 PM.
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