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

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VP
Joined: 22 Nov 2007
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4 couples wish to stand in a row for a group of photo. How [#permalink]

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16 Dec 2007, 01:11
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 of photo. How many arrangements of the 8 people are possibile if each person must stand next to his or her partner?

A. 324
B. 352
C. 384
D. 426
E. 512
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 544

Kudos [?]: 3550 [0], given: 360

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16 Dec 2007, 01:18
C

N=4!*(2C1)^4=24*16=384
VP
Joined: 22 Nov 2007
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ok [#permalink]

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16 Dec 2007, 02:10
yes...good...see my next one
Intern
Joined: 13 Jun 2007
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16 Dec 2007, 05:56
"Walker" i agree with the answer but what is the logic you used?
For example i came to the same answer by

4P4 give all the SETS of 4 couple arrangements
2P2 gives the arrangement within a couple
2P2^4 gives the total arrangement within one set

4P4 * 2P2^4

Why did you use 2C1?
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 544

Kudos [?]: 3550 [0], given: 360

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16 Dec 2007, 06:33
alexperi wrote:
Why did you use 2C1?

4!=4P4 as your way.
2C1=2 means two different cases for first position in a couple.

I think your way is better one in general. Because nPn is more appropriate in the case of subsets with more than two members.

nPn=nC1*(n-1)C1*...*1C1 - is general relation between our ways
16 Dec 2007, 06:33
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4 couples wish to stand in a row for a group of photo. How

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