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Math Help - question on permutation

  1. #1
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    question on permutation

    Two Mathematics books, three Chemistry books and four Physics books are to be arranged in a row on a shelf. Given that all the books are different, find the number of ways of arranging the books if no two Physics books are placed together.

    the answer should be 43200.
    I am able to find out the number of possible arrangement for the 4 physics books so that no two of them are placed together, which is 15.
    so 15x4!x5!=43200.
    But this method is a bit stupid since I need to count the possible arrangement one by one.

    Is there a simpler way to solve it??? Please help!!
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  2. #2
    Grand Panjandrum
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    Don't post the same question to multople areas on the MathHelpForum.

    It confuses the helper and makes work for the Moderators.

    RonL
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  3. #3
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    ooops...sorry. I post it at the wrong section in the first place
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  4. #4
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    Hello, acc100jt!

    I can't locate your duplicate post, so this may be a repeat of someone else's solution.


    Two Math books, 3 Chemistry books and 4 Physics books are to be arranged in a row on a shelf.
    Given that all the books are different, find the number of ways of arranging the books
    if no two Physics books are placed together.

    The answer should be 43,200.

    Consider the 2 Math books and 3 Chemistry books placed in a row
    . . with a space between them and at the ends.

    . . \_\_\:X\;\_\_\;X\;\_\_\;X\;\_\_\;X \;\_\_\;X\;\_\_

    These five books can be arranged in 5! = 120 ways.


    Now place the 4 Physics books.
    We have a choice of six spaces and we will choose four of them.
    There are: . P(6,4) = \frac{6!}{2!} = 360 ways.

    Therefore, there are: . 120 \times 360 \:=\:43,200 possible arrangements.

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    Thanks a lot!! soroban!!!!
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