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Homework Help: Probability arranging books

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    there are 7 different french books and 7 different Spanish books, how many ways are there to arrange them on a shelf
    a. books of the same language must be group together, French on left and Spanish on Right?
    b. French and Spanish books must alternate in the grouping, beginning with a French book?

    I tried doing 7!x7! for both of them but i don't think i am right?

    I have no idea how to approach this
     
  2. jcsd
  3. Nov 19, 2012 #2

    haruspex

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    Sounds right to me.
     
  4. Nov 22, 2012 #3
    for the first one 7!x7! seems right, but for the 2nd one I think (not sure...!!!) it's :

    7C1X7C1 X 6C1X6C1 X 5C1X5C1 X 4C1X4C1 X 3C1X3C1 X 2C1X2C1 X 1C1X1C1
     
  5. Nov 22, 2012 #4

    CAF123

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    I think an argument could go: there are 14 choices for the first book, (French or Spanish). There are then 7 choices for the next book (If first was French, this one must be Spanish), then 6 choices for next, (has to be French), then 6 choices for next (has to be Spanish)....and so on. In this Q, the French book is first so what you wrote is correct.
     
    Last edited: Nov 22, 2012
  6. Nov 22, 2012 #5

    Ray Vickson

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    Another argument: for each arrangement of the French books, leave a space between successive books and fill those spaces with the Spanish books, one book per space.

    RGV
     
    Last edited: Nov 22, 2012
  7. Nov 22, 2012 #6

    haruspex

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    How is that different from 7!x7!?
    The two obviously have the same answer. Either way, there is a fixed set of 7 positions that can be taken by the French books, and another fixed set of 7 that can be taken by the Spanish, independently.
     
  8. Nov 22, 2012 #7

    Ray Vickson

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    It's not different; it's just another argument that the OP may, or may not, prefer.

    RGV
     
  9. Nov 22, 2012 #8

    haruspex

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    I was replying to MadAtom, who wrote:
    the first one 7!x7! seems right, but for the 2nd one I think (not sure...!!!) it's :​

    Seems to me MadAtom implied 7!x7! was wrong for the second question.
     
  10. Nov 23, 2012 #9
    I thought so, but the result is the same... sorry.
     
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