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Two couples and a single person must be randomly seated in a

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Two couples and a single person must be randomly seated in a [#permalink] New post 09 Oct 2005, 08:26
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100% (03:53) correct 0% (00:00) wrong based on 2 sessions
Two couples and a single person must be randomly seated in a row of 5 chairs. What is the probability that neither couples sit together in adjacent chairs?
1) 1/5
2) 1/4
3) 3/8
4) 2/5
5) 1/2

OA to follow.
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 [#permalink] New post 09 Oct 2005, 09:38
fresinha12 wrote:
is it 2/5?

will explain if correct.


Yup, OA is 2/5... explain please. Thx.
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 [#permalink] New post 09 Oct 2005, 15:46
I got 2/5 ... Solved by brute force.

fresinha12: Could you post the faster way?
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 [#permalink] New post 09 Oct 2005, 18:22
Hi,
I got 2/5 as well and i followed this method. Please let me know if this seems ok to you.
Prob ( both couples not sitting together ) =
1 - ((probability that exactly one couple sits together) + ( probability that both couples sit together))

Total number of ways of arranging 5 people without any restriction - 5! ( this is also the total number of possibilities)

Number of Ways in which 1 couple sit together but the others sit in any order - 4! * 2 ( If one couple has to sit together consider, consider the couple sitting together as one unit. so we have 4 units to arange in 4! ways, within their unit, the couple can sit in 2 ways, hence 4!*2)

Similarly number of ways in which both couples sit together = 3!* 4 ( multiplication by 4 because of 2 couples involved with 2 arrangements each)

So finally we get (5! - ((3!*4)+(4!*2)) ) / 5! = 2/5.
  [#permalink] 09 Oct 2005, 18:22
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