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# 4 couples wish to stand in a row for a group photo. How many

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VP
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4 couples wish to stand in a row for a group photo. How many [#permalink]

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02 Jan 2008, 09:47
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

4 couples wish to stand in a row for a group photo. How many arrangements of the 8 people are possible if each person must stand next to his or her partner?

A. 324
B. 352
C. 384
D. 426
E. 512
VP
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02 Jan 2008, 10:02
bmwhype2 wrote:
4! 2! 2! 2! 2! = 384

yeah ... it could also be written as 4!*2^4 which means...4! are the positions that all couples may have globally; 2^4 is 2^n and stands for the positions all persons can have in each couple...

can someone explain me the difference with 7-t56557
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02 Jan 2008, 10:21
marcodonzelli wrote:
bmwhype2 wrote:
4! 2! 2! 2! 2! = 384

yeah ... it could also be written as 4!*2^4 which means...4! are the positions that all couples may have globally; 2^4 is 2^n and stands for the positions all persons can have in each couple...

can someone explain me the difference with 7-t56557

there is a sharp difference in the questions.
the one i posted is choosing 4 people into a group, in which we want no pairs.
In your question, you are simply arranging couples.
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02 Jan 2008, 10:24
bmwhype2 wrote:
marcodonzelli wrote:
bmwhype2 wrote:
4! 2! 2! 2! 2! = 384

yeah ... it could also be written as 4!*2^4 which means...4! are the positions that all couples may have globally; 2^4 is 2^n and stands for the positions all persons can have in each couple...

can someone explain me the difference with 7-t56557

there is a sharp difference in the questions.
the one i posted is choosing 4 people into a group, in which we want no pairs.
In your question, you are simply arranging couples.

yeah, I see...but why do we have to divide for 4! in your question? I face this kind of problems in combinatorics...
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02 Jan 2008, 10:36
i think you are getting permutations and combinations confused.

formula for combinations is nCr where n!/r! (n-r)!
that is where the 4 comes from.
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02 Jan 2008, 10:38
http://www.themathpage.com/aPreCalc/per ... s.htm#perm
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02 Jan 2008, 11:04
For this question, why is it wrong to think of it this way:

each couple can be put in one of 4 positions, so 4! for each of 4 couples. But, each couple can also be arranged in 2! ways.

Therfore, for each couple, the number of ways is 4!2! = 48 ways

For four couples, 48 x 4 = 192 ways
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02 Jan 2008, 11:15
pmenon wrote:
For this question, why is it wrong to think of it this way:

each couple can be put in one of 4 positions, so 4! for each of 4 couples. But, each couple can also be arranged in 2! ways.

Therfore, for each couple, the number of ways is 4!2! = 48 ways

For four couples, 48 x 4 = 192 ways

because 4!*2! is different from 4*2!
arranging 4 elements vs multiplying by a constant of 4
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02 Jan 2008, 12:44
bmwhype2 wrote:
4! 2! 2! 2! 2! = 384

Nice job!
Some hieroglyphs: 4P4*(2P2)^4
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02 Jan 2008, 23:01
I didn't got the point with superscript, how did you come up with 4 in the power?
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03 Jan 2008, 00:58
kazakhb wrote:
I didn't got the point with superscript, how did you come up with 4 in the power?

4P4 - arrangements of 4 couples in a raw
2P2 - arrangements of 2 people in a couple.
(2P2)^4 - arrangements of 2 people within a couple for all 4 couples .
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Re: focusing on combinatorics   [#permalink] 03 Jan 2008, 00:58
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