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May 16th, 2010, 09:57 PM   #1
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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
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May 16th, 2010, 11:38 PM   #2
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Re: Permutations

Yes, I should have realized that the people would be unique. Good job mrtwhs.

Sorry about that.
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May 17th, 2010, 02:35 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
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May 17th, 2010, 07:34 AM   #4
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Re: Permutations

Hello, mrtwhs!

I like your solutions!

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