Robert must read a few books from his home library. He read 4 out of 6 books from the top shelf, then 2 out of 3 books from the middle shelf and then 3 out of 6 books from the bottom shelf. In how many ways can Robert read the books, if different orders in which the books will be read count as different ways?

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asked by yaz
  1. First let's choose the books
    top shelf = C(6,4) = 15
    middle shelf = C(3,2) = 3
    bottom shelf = C(6,3) = 20

    so he can read 15x3x20 or 900 different combinations of 9 books

    Now for each of the choices of 9 books, he can read them in 9! ways.

    so the number of ways he can read the books
    = 900(9!)
    = 326,592,000

    2nd interpretation:
    He has a choice of 15 different groups of 4 books from the top shelf
    these can be read in 15(4!) or 360 orders.

    He has choice of 3 pairs of books from the middle, these can be read in 3(2!) or 6 ways
    He has a choice of 30 triples of books from the bottom, which he can read in 30(3!) or 180
    So the number of arrangement if he must follow the order top, middle bottom books
    = 360(6)(180)
    = 388,800

    I would go with my first solution

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    posted by Reiny
  2. Reiny is partially correct. However, instead of using combinations you would use permuations. So it would be P(6,4)*P(3,2)*P(6,3), and that would leave you with the answer 259200.

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    posted by Nathan

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