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

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Manager
Joined: 09 Aug 2009
Posts: 50

Kudos [?]: 10 [0], given: 1

Re: couples in a row [#permalink]

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27 Sep 2009, 23:48
I also kinda need the same answer.....[/quote]

I got it..
To get a feel for it
COND1: Just write all the possible condition when the couple sits together
COND2: Write down the possible outcome for only one couple sits together.
You can find repeated sittinf arrangement in COND2 which consist of all the COND1 outcome.

Hope this helps you

/Prabu

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Intern
Joined: 29 Sep 2009
Posts: 6

Kudos [?]: 4 [0], given: 0

Schools: ISB, Wharton
Re: couples in a row [#permalink]

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10 Oct 2009, 21:35
Thank you very much!!
_________________

Thanks,
Just think differently, there is a easier solution:)

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Manager
Joined: 09 Aug 2010
Posts: 106

Kudos [?]: 57 [0], given: 7

Re: couples in a row [#permalink]

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09 Apr 2011, 07:27
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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VP
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Re: couples in a row [#permalink]

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01 May 2011, 23:28
Interesting solution.
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VP
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Kudos [?]: 648 [0], given: 70

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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, 20:05
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..................

Kudos [?]: 648 [0], given: 70

Math Expert
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Kudos [?]: 129091 [0], given: 12194

Re: Two couples and one single person are seated at random in a [#permalink]

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06 Oct 2012, 05:36
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
_________________

Kudos [?]: 129091 [0], given: 12194

Re: Two couples and one single person are seated at random in a   [#permalink] 06 Oct 2012, 05:36

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