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Director
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Counting [#permalink]

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New post 20 Aug 2010, 15:58
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3 couples need to be arranged for a group picture , and the couples cannot be separated how many arrangements is possible
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Director
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New post 20 Aug 2010, 16:36
relovchkin made me think again, i always have confusion as to whether 2^3 or 2! x 3.. in this case 3! is clear and then my thinking was that i can rearrange each couple in 2 ways, then there are 3 couples so it is 3! x 2 x 2 x 2.. When would 3! x 2! x 3 be right?
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New post 21 Aug 2010, 04:19
First consider the 3 couples each as single unit. Hence 3 couples can be arranged in 3! ways. And then for each of those 6 combination there are 8 ways in which individuals in a couple can be arranged.
Hence 6*8 = 48 ways.
Eg:
A,B,C are 3 couples.
6 ways to arrange them is:
ABC BCA CBA
ACB BAC CAB
Now if a1a2, b1b2, c1c2 are the individuals, then they can be arranged in 8 ways:
a1a2,b1b2,c1c2
a2a1,b1b2,c1c2
a1a2,b2b1,c1c2
a2a1,b2b1,c1c2
a1a2,b1b2,c2c1
a2a1,b1b2,c2c1
a1a2,b2b1,c2c1
a2a1,b2b1,c2c1
Similar arrangements are possible for the rest of the 5 couple arrangements.

Am I right in my explanation???
Re: Counting   [#permalink] 21 Aug 2010, 04:19
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