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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
Posts: 555

Kudos [?]: 216 [0], given: 0

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

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23 Jul 2008, 01:38
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

Kudos [?]: 216 [0], given: 0

Current Student
Joined: 12 Jun 2008
Posts: 287

Kudos [?]: 58 [0], given: 0

Schools: INSEAD Class of July '10

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23 Jul 2008, 02:28
First you have to pick which of the 3 couples you will take your people from : $$C_4^3 = 4$$ choices

Then you are left with 3 couples and you can pick for each one any of the 2 persons: $$2^3 = 8$$ possibilities

Total number of possible groups = $$4*8 = 32$$

Kudos [?]: 58 [0], given: 0

Director
Joined: 27 May 2008
Posts: 541

Kudos [?]: 354 [0], given: 0

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23 Jul 2008, 02:50
total people = 4*2 = 8

total combination = 8C3 = 56

P (selecting first person) = 8/8 ( can be any one)
P (selecting second person) = 6/7 ( cant take spouse of first)
P (selecting third person) = 4/6 (cant take spouses of first two)

P = 8/8 * 6/7 * 4/6 = 4/7

fav ways = 4/7 * 56 = 32 ............. answer

Kudos [?]: 354 [0], given: 0

SVP
Joined: 28 Dec 2005
Posts: 1549

Kudos [?]: 173 [0], given: 2

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23 Jul 2008, 04:50
32 for me as well.

for the first spot, we can pick any one person out of the 8, i.e. 8C1 = 8

for the second spot, we need to pick one person out of the remaining 6 ... not the remaining 7, since we dont want to pick the spouse of the person we selected first ... so 6C1=6

same logic for the third spot to get 4C1=4

finally, we have to divide out by 3!, since this represents the different permutations of the three spots, and we dont care about order so we end up with (8x6x4)/6 = 32

Kudos [?]: 173 [0], given: 2

Re: coupled problem!!!   [#permalink] 23 Jul 2008, 04:50
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