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

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Director
Joined: 05 May 2004
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Location: San Jose, CA
Two couples and one single person are seated at random in a [#permalink]

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19 Sep 2004, 22:14
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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?

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Director
Joined: 20 Jul 2004
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20 Sep 2004, 06:27
Pb that neither couple sit together
= 1 - Pb that couples always sit together
= 1 - 3!/5!
= 1 - 1/20
= 19/20

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Senior Manager
Joined: 06 Dec 2003
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Location: India

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20 Sep 2004, 13:17
total no of arrangement - 5! / 2! * 2! = 30
Both couple sit together - 3! = 6

Couple - 1 sits together + Both couple sit together = 4! / 2! = 6
Couple - 2 sits together + Both couple sit together = 4! / 2! = 6

so no of cases, when neither couple will sit together = 30 - (6 + 6 + 6) = 12
So probability = 12/30 = 2/5

whats the OA, dear ?

Dharmin
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Perseverance, Hard Work and self confidence

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20 Sep 2004, 14:56
I have 4/5
Total possible outcomes: 5!
unfavorable outcomes with couple XX, YY sitting together:
(XX)(YY)(Z)
Since Xs and Ys are interchangeable, we have 3!*2!*2!
1 - (3!*2!*2!)/5! = 96/120 = 4/5
hardworker, I think you forgot to add 2!*2! for when X's and Y's are interchanged
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Best Regards,

Paul

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Senior Manager
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20 Sep 2004, 15:23
Paul, you have calculated the probability of 1-(BOTH are together).
There is also a possibility that only one of the couples sits together.

total = 5! = 120.
both together = 3!*2!*2! = 24.
one together = 4!*2 - 24 = 24

Last edited by Dookie on 20 Sep 2004, 15:47, edited 1 time in total.

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GMAT Club Legend
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20 Sep 2004, 15:39
True, true... Dharmin was right
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Best Regards,

Paul

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Director
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20 Sep 2004, 17:54
Thats why I shouldn't answer at 5:30 AM. Good question though.

Total number of cases = 5! = 120
Number of cases both couples sit together = 3! X 2 X 2 = 24
Number of cases first couple sits together = 4! X 2 = 48 - 24 = 24
Number of cases second couple sits together = 4! X 2 = 48 - 24 = 24

Pb that both couples do not sit together
= 1 - Pb that both sit together - Pb that first couple sits together - Pb that second couple sits together
= 1 - (24+24+24)/120
= 1 - 3/5
2/5

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Intern
Joined: 02 Aug 2004
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21 Sep 2004, 16:40
Thanks for the heads up. I'll be more careful when i post i guess.

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Senior Manager
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21 Sep 2004, 18:08
Consider the placement of the single person:

Single in Chair 1:
4 possible for chair 2
2 possible for chair 1

8 arrangements
(same as if single in Chair 5)

Single in Chair 2:
4 possible for chair 1
2 possible for chair 3

8 Arrangements
(same as if single in Chair 4)

Single in the Middle
4 possible in chair 1
2 possible in chair 2
2 possible in chair 4
16 arangements

probablility = 8 + 8 + 16 + 8 + 8 / 5!
= 48/120 = 2/5

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21 Sep 2004, 18:08
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