December 6th, 2012, 12:29 AM  #1 
Senior Member Joined: Mar 2012 Posts: 118 Thanks: 0  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?

December 6th, 2012, 12:39 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,710 Thanks: 675  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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