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Counting Problem

  1. Apr 12, 2005 #1
    Hi, I wasn't sure how to approach this problem:

    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.

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

    Help please! thanks.
     
  2. jcsd
  3. Apr 12, 2005 #2

    Hurkyl

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    How many ways can you place the green one, then place the rest?

    P.S. please don't multiple post.
     
  4. Apr 12, 2005 #3
    um 3? lol I don't get it...
     
  5. Apr 12, 2005 #4

    HallsofIvy

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    The DO it.

    Suppose you place the green stone in the second place. How many different ways are there to place the other 4 stones?

    Suppose you place the green stone in the third place. How many different ways are there to place the other 4 stones?

    Suppose you place the green stone in the fourth place. How many different ways are there to place the other 4 stones?

    Okay, now how many ways is that all together?
     
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