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Math Help - arrange book's question

  1. #1
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    Cool arrange book's question

    places on at will the bookshelf 10 books, asks 5 books which in which assigns to put on the same place probability
    Last edited by neworld222; June 29th 2007 at 05:24 AM.
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  2. #2
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    What is that supposed to mean?. I am sorry, but it makes no sense.
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  3. #3
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    Hello, neworld222!

    I think it's Yoda, the Jedi Master . . .


    places on at will the bookshelf 10 books,
    asks 5 books which in which assigns to put on the same place probability
    I'll take a wild guess . . .

    Ten distinct books are placed randomly on a shelf (in a row).
    What is the probability that a particular five books are adjacent?


    There are 10! possible arrangements.

    Duct-tape those five books together.
    Then we have six "books" to arrange: . \left\{\boxed{ABCDE},\,F,\,G,\,H,\,I,\,J\right\}
    . . They can be arranged in 6! ways.

    For each of these, the five books can be permuted in 5! ways.
    . . Hence, there are: . (6!)(5!) desirable arrangements.

    The probability is: . \frac{(6!)(5!)}{10!} \;=\;\frac{1}{42}

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  4. #4
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    Thumbs up

    Quote Originally Posted by Soroban View Post
    Hello, neworld222!

    I think it's Yoda, the Jedi Master . . .


    I'll take a wild guess . . .

    Ten distinct books are placed randomly on a shelf (in a row).
    What is the probability that a particular five books are adjacent?


    There are 10! possible arrangements.

    Duct-tape those five books together.
    Then we have six "books" to arrange: . \left\{\boxed{ABCDE},\,F,\,G,\,H,\,I,\,J\right\}
    . . They can be arranged in 6! ways.

    For each of these, the five books can be permuted in 5! ways.
    . . Hence, there are: . (6!)(5!) desirable arrangements.

    The probability is: . \frac{(6!)(5!)}{10!} \;=\;\frac{1}{42}

    yes ,you give what i need .English is not my first lanuage,so i sorry for that i made you take a wild guess
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