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# Four couples are to select a group that consists of three

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Current Student
Joined: 11 May 2008
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Four couples are to select a group that consists of three [#permalink]

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02 Sep 2008, 20:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

. Four couples are to select a group that consists of three people from different couples. How many such groups are possible?
(A) 24
(B) 28
(C) 32
(D) 36
(E) 40

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Senior Manager
Joined: 09 Oct 2007
Posts: 463

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02 Sep 2008, 20:18
My take: (8 x 6 x 4)/3! = 32 = C

What's the OA?

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SVP
Joined: 07 Nov 2007
Posts: 1791

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Location: New York

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02 Sep 2008, 20:19
arjtryarjtry wrote:
. Four couples are to select a group that consists of three people from different couples. How many such groups are possible?
(A) 24
(B) 28
(C) 32
(D) 36
(E) 40

= 8C1 * 6C1*4C1 / 3!
= 32
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Current Student
Joined: 11 May 2008
Posts: 554

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02 Sep 2008, 20:24
ans is 32. but both pls provide expln as to how you came at the ans.
for eg . why is it 8c1 and not 8c3 etc etc.
the detailed working is preferred

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VP
Joined: 05 Jul 2008
Posts: 1402

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02 Sep 2008, 20:28
Total number of ways 3 people can be picked from 8 is 8 c 3 = 56

need to find possibilities in such a way that no couple is on the team

Lets find the possibilities that a couple is on the team. Any of the 4 couples can be on the team. One other person can be any one from the remaining 6 So total number of such possibilities = 4X6 =24

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SVP
Joined: 07 Nov 2007
Posts: 1791

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Location: New York

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02 Sep 2008, 21:15
arjtryarjtry wrote:
ans is 32. but both pls provide expln as to how you came at the ans.
for eg . why is it 8c1 and not 8c3 etc etc.
the detailed working is preferred

F1M1
F2M2
F3M3
F4M4

= Select any person from 8 members(8C1 ways) * select another person (exclude the spouse of previous person) 6C1 * select third person from the remaining(exclude spouses of first and second persons) 4C1 / 3!

F1F2F3 -- can be arranged in 3! ways
F1F2F3 - F2F1F3 - F1F3F2 - F2F3F1 -F3F1F2 - F3F2F1 .. (all these are same)
So to avoid duplicates divide by 3!.

is it clear now.
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Intern
Joined: 02 Sep 2008
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03 Sep 2008, 11:55
Yep,

I am also getting 32 as answer.

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Re: combination   [#permalink] 03 Sep 2008, 11:55
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# Four couples are to select a group that consists of three

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