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December 6th, 2012, 12:29 AM   #1
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permutation

in how many ways can 3 math I books,5 math II books 4 math III books and 2 math IV books be arranged on a shelf?
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December 6th, 2012, 12:39 PM   #2
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Re: permutation

Assuming books at the same level are identical, then you have 14! not using identical, divided by 3!5!4!2! permuting identical books. Net = 14!/(3!5!4!2!).
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