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Math Help - Colored Stones

  1. #1
    Bertha
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    Colored Stones

    You have 5 differently colored stones-red, orange, blue, green, purple. If the green stone cannot be placed at the front or the back of the sequence, how many possible arrangements can you make?

    I know that without the above restriction, the amount would be 5!=120.

    But I don't get how to use the restriction.

    The back of the book says that the answer is 72.

    Thanks for any help
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  2. #2
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    That would be 5! - 2*4!
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  3. #3
    Site Founder MathMan's Avatar
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    Quote Originally Posted by paultwang
    That would be 5! - 2*4!
    Can you please explain I don't understand the logic behind that.

    Thanks
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  4. #4
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    When green is placed at the back:
    ROBPG, ROPBG, ..... (4! combinations)

    When green is placed at the front:
    GROBP, GROPB, ..... (4! combinations)

    5! - 2*4!
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  5. #5
    MHF Contributor

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    There are 5 stones. Since you CANNOT place the green stone at the front, there are 4 stones you CAN place there. Assuming you have placed that stone, there are 4 stones left. Since you CANNOT place the green stone at the end (and one of the 4 left is green), there are 3 stones left that you CAN place there. Now you have 3 stones left to put in the remaining 3 places- of course there are 3! ways of doing that.
    The total number of ways you can place the 5 stones WITHOUT the green stone being first or last, is (4)(3)(3!)= 4*3*6= 72. That is, of course, exactly the same as 5!- 2(4!).
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