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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
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

4 couples wish to stand in a row for a group of photo. How [#permalink]  16 Dec 2007, 01:11
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: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2149 [0], given: 359

Expert's post
C

N=4!*(2C1)^4=24*16=384
VP
Joined: 22 Nov 2007
Posts: 1105
Followers: 7

Kudos [?]: 193 [0], given: 0

ok [#permalink]  16 Dec 2007, 02:10
yes...good...see my next one
Intern
Joined: 13 Jun 2007
Posts: 48
Followers: 1

Kudos [?]: 5 [0], given: 0

"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: 3578
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 407

Kudos [?]: 2149 [0], given: 359

Expert's post
alexperi wrote:
Why did you use 2C1?

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
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