text: Four couples are to select a group that consists of three : Quant Question Archive [LOCKED]
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  # Four couples are to select a group that consists of three
  
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  Four couples are to select a group that consists of three [#permalink]
  
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  23 Jul 2008, 00:38
  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
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  23 Jul 2008, 01:28
  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 = $$4*8 = 32$$
  
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  23 Jul 2008, 01:50
  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
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  23 Jul 2008, 03:50
  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
  Re: coupled problem!!!   [#permalink] 23 Jul 2008, 03:50
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