May 16th, 2010, 10:57 PM  #1 
Newbie Joined: Apr 2010 Posts: 24 Thanks: 0  Permutations
find the number of ways of arranging: 3 dancers, 2 singers and 4 students if a) each group must stand together b) no students can stand next to each other c) the oldest from each group must stand in the middle 3 positions if someone could help me out 
May 17th, 2010, 12:38 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,742 Thanks: 1000 Math Focus: Elementary mathematics and beyond  Re: Permutations
Yes, I should have realized that the people would be unique. Good job mrtwhs. Sorry about that. 
May 17th, 2010, 03:35 AM  #3  
Senior Member Joined: Feb 2010 Posts: 636 Thanks: 106  Re: Permutations Quote:
a) line up the dancers > 3! line up the singers > 2! line up the students > 4! line up the groups > 3! so the answer is (3!)(2!)(4!)(3!) = 1728 b) line up the dancers and singers > 5! spread them out with some spaces _X_X_X_X_X_ there are now six spaces into which you could put the four students > 6P4 so the answer is (5!)(6P4) = (120)(360) = 43200 c) line up the three oldest > 3! line up the other six > 6! stuff the three into the middle so the answer is (6!)(3!) = 4320  
May 17th, 2010, 08:34 AM  #4 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Permutations Hello, mrtwhs! I like your solutions! 

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