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# There are 5 married couples and a group of three is to be

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Senior Manager
Joined: 19 Oct 2010
Posts: 252

Kudos [?]: 91 [0], given: 27

Location: India
GMAT 1: 560 Q36 V31
GPA: 3
There are 5 married couples and a group of three is to be [#permalink]

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29 Jan 2011, 14:19
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I don't understand the answer. In fact, I think it's wrong.
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petrifiedbutstanding

Kudos [?]: 91 [0], given: 27

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7736

Kudos [?]: 17766 [3], given: 235

Location: Pune, India

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29 Jan 2011, 20:32
3
KUDOS
Expert's post
petrifiedbutstanding wrote:
There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?

I don't understand the answer. In fact, I think it's wrong.

Another way to approach this problem would be to use basic counting principles.

10 Available people

3 Available positions _ _ _
In the first position, you can put any one of the 10 people. In the second position, you can put any one of 8 people (not 9 because the spouse of the person put in 1st position cannot be put in the second position). In the third position you can put any one of 6 people (2 people put in previous places and their spouses cannot be put in the third position)

Total number of arrangement = 10*8*6
But mind you, these are arrangements. We just need a group of 3 people. They don't need to be arranged in the group. So we divide these arrangements by 3! to get just the combination.

(I do hope I have read the question properly this time petrifiedbutstanding!)
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Karishma
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Kudos [?]: 17766 [3], given: 235

Director
Joined: 01 Feb 2011
Posts: 726

Kudos [?]: 146 [0], given: 42

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22 Apr 2011, 14:35
My answer is inline with Karishma's

10*8*6 / 3! = 80

Kudos [?]: 146 [0], given: 42

Re: Combination/Permutation problem, couples   [#permalink] 22 Apr 2011, 14:35
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