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Thread: Disorder problem

  1. #1
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    Disorder problem

    Hi,
    n people hang their coats and hats in a bar.Then ,each one choose randomly a hat and a coat.


    What is the probability that:

    1)None of them will take his one hat or his own hat?

    2)None of them will take both his coat and his hat?

    Thank's in advance
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  2. #2
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    Re: Disorder problem

    1) look up Derrangement numbers. The answer is $\dfrac{!n}{n!}=\dfrac 1 e$

    2) $\dfrac{1}{e^2}$
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  3. #3
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    Re: Disorder problem

    Quote Originally Posted by hedi View Post
    Hi,
    n people hang their coats and hats in a bar.Then ,each one choose randomly a hat and a coat.
    What is the probability that:
    1)None of them will take his one hat or his own hat?
    2)None of them will take both his coat and his hat?
    Please review the statement. It is vague. Are the hat and coat hanged together?
    Or are they put on separate racks (the question makes more sense if they are)?
    This is a derangement problem.
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  4. #4
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    Re: Disorder problem

    can you elaborate your answer,please?
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    Re: Disorder problem

    Quote Originally Posted by hedi View Post
    can you elaborate your answer,please?
    @ Hedi, Did you read the references that we provided?
    The number of ways $n$ objects can be rearranged so that no one is in its correct position is $\displaystyle\mathscr{D}_n= n!\sum\limits_{k = 0}^n {\dfrac{{{{( - 1)}^k}}}{{k!}}}$
    SEE HERE $\mathscr{D}_6=265$ which means if the string $123456$ is randomly rearranged there are $265$ rearrangements in which no one of those six digits is in its correct position.
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  6. #6
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    Re: Disorder problem

    The first question is ment to be "nobody will get his own coat or his own hat"
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  7. #7
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    Re: Disorder problem

    Quote Originally Posted by hedi View Post
    The first question is meant to be "nobody will get his own coat or his own hat"
    You don't answer our sincere questions and moreover you don't think we read English.
    I doubt that you can write clear English. So we are left to assume that these men pick a coat at random and then pick a hat at random.
    Now because $\sum\limits_{k = 0}^\infty {\dfrac{{{{( - 1)}^k}}}{{k!}}} = \dfrac{1}{e}$
    We can use that to model $\mathscr{D}_n=\left\lfloor {\frac{{n!}}{e} + \frac{1}{2}} \right\rfloor \text{ for }n\ge 3$.
    Look at this table: SEE HERE

    Let $C$ be the event that none of the $n$ men gets his own Coat. Let $H$ be the event that none of the $n$ men gets his own Hat.
    $\mathcal{P}(C\cup H)=\mathcal{P}(C)+\mathcal{P}(H)-\mathcal{P}(C\cap H)$ i.e. Does not get his coat or hat.

    Now if you were not so self-important you may have done us the honor of explaining the setup of the question.
    If there were two different racks, one for coats the other for hats, then clearly the selections would be independent.

    So if we assume Independence how do you finish? DON'T ASK!
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