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# Two couples and one single person are seated at random in a

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Intern
Joined: 29 Sep 2009
Posts: 6
Schools: ISB, Wharton
Re: couples in a row  [#permalink]

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10 Oct 2009, 20:35
Thank you very much!!
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Thanks,
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Senior Manager
Joined: 22 Dec 2009
Posts: 301
Re: couples in a row  [#permalink]

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14 Feb 2010, 11:03
1
marcodonzelli wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

Choices are 1/5, 1/4 , 3/8, 2/5 and 1/2

Probability both couples sit together = 3! x 2! x 2! = 24

Probability that only first couple sits together = 4! x 2! - both couple sits together = 48 - 24 = 24
Probability that only second couple sits together = 4! x 2! - both couple sits together = 48 - 24 = 24

Therefore no couple sits together =1 - (24 x 3) / 5! = 1 - 3/5 = 2/5
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Manager
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Posts: 91
Re: couples in a row  [#permalink]

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09 Apr 2011, 06:27
took me almost forever to get why you have 16 for this _ _ 5 _ _, whew!

eschn3am wrote:
marcodonzelli wrote:
Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?

Choices are 1/5, 1/4 , 3/8, 2/5 and 1/2

There are 5! = 120 ways to seat the 5 people.

I'll use the same naming of the people. (1,2) (3,4) and 5 as the single.

5 _ _ _ _ = _ _ _ _ 5 Now how many ways can we sit people without couples touching?

4*2*1*1 = 8

and since 5 _ _ _ _ = _ _ _ _ 5 we have 8*2 = 16 ways to seat people without couples touching when the single is on the end.

_ 5 _ _ _ = _ _ _ 5 _ Now how many ways can we sit people without couples touching?

4*2*1*1 = 8

and since we have two seating arranges for that to work 8*2 = 16 ways to seat people when the single is sitting one spot from the end.

_ _ 5 _ _ with no mirror image

4*2*2*1 = 16

16+16+16 = 48

48/120 = 2/5

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Re: couples in a row  [#permalink]

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01 May 2011, 22:28
Interesting solution.
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Re: Two couples and one single person are seated at random in a  [#permalink]

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05 Oct 2012, 19:05
Need help
[A,A1,B,B1],C
Prob of couples together= (4!*2!)/5!=2/5
Prob That Couples Not together = 1-2/5 =3/5
Pls point out what am i not accounting..................
Math Expert
Joined: 02 Sep 2009
Posts: 52387
Re: Two couples and one single person are seated at random in a  [#permalink]

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06 Oct 2012, 04:36
Archit143 wrote:
Need help
[A,A1,B,B1],C
Prob of couples together= (4!*2!)/5!=2/5
Prob That Couples Not together = 1-2/5 =3/5
Pls point out what am i not accounting..................

OPEN DISCUSSION OF THIS QUESTION IS HERE: two-couples-and-one-single-person-are-seated-at-random-in-a-92400.html
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Posts: 9463
Re: Two couples and one single person are seated at random in a  [#permalink]

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30 Jul 2018, 11:00
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Two couples and one single person are seated at random in a &nbs [#permalink] 30 Jul 2018, 11:00

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