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Math Help - derangement question

  1. #1
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    Exclamation derangement question

    A brand new standard deck of 52 playing cards comes with the cards arranged in a particular order. Before using them, the cards are shuffled. What is the probability that exactly 10 of the cards will remain in their original position?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Clearly it's equal to the number of possible derangements of the remaining 42 cards, divided by 52!. The number of derangements of n objects is the integer closest to n!/e, in this case 51687286544138714389965559095310738479145874768856  1. So the probability is


    \frac{51687286544138714389965559095310738479145874  7688561}{80658175170943878571660636856403766975289  505440883277824000000000000}.
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  3. #3
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    Quote Originally Posted by Bruno J. View Post
    Clearly it's equal to the number of possible derangements of the remaining 42 cards, divided by 52!. The number of derangements of n objects is the integer closest to n!/e, in this case 51687286544138714389965559095310738479145874768856  1. So the probability is


    \frac{51687286544138714389965559095310738479145874  7688561}{80658175170943878571660636856403766975289  505440883277824000000000000}.
    Don't forget to take into account the possible choices of the 10 cards that remain in their original position.
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Oh yeah, where is my head! That'll teach me to try helping people at 2:00AM.
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