My Math Forum Permutations

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 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
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Re: Permutations

Quote:
 Originally Posted by football 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
I'm going to assume the 9 people are unique (!)

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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