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Manager
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Two Couples and One Single Person are seated at random in a [#permalink]
 03 Aug 2005, 06:31
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 adjacents chairs?
(A) 1/5
(B) 2/5
(C) 4/5
(D) 5/7
(E) 3/4
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Manager
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C
ways together:
randomly = 4! = 24 and
a couple next to a couple like in x1x2y1y2... = 16
total together = 8+16 = 24
Pr not together = 1-together = 1 - 24/5! = 4/5.
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Senior Manager
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Dan wrote: C
ways together:
randomly = 4! = 24 and
a couple next to a couple like in x1x2y1y2... = 16
total together = 8+16 = 24
Pr not together = 1-together = 1 - 24/5! = 4/5.
plz dan why do you divide 24 by 5! thanks
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Senior Manager
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A
I felt this is equivalent to ask what is the probability of having
the single sitting in the middle seat. (that is the only situation
where the two couples are not sitting next to each other)
so that is 1 in 5, thus 1/5
the couple together's probability should be quite small,
4/5 seems to be too high.
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Manager
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A also . Same reason as qpoo.
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Manager
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I go with A.
The total outcomes are 5!
The number of the event that there is a couple sitting adjacently each other is 6*4*2*2=96.
Considering that:
The five people are A,A', B,B',C with AA' are a couple and BB' are the other couple. The five seats are 1,2,3,4,5.
We can see if AA' sitting in two adjacent seats, there are six ways (6) to arrange the remainders in the other three seats.
Moreover, there are four ways (4) to seat A and A' in five seats so that A and A' sit adjacently.
Not only that, there are two ways (2) to seat A and A' adjacently.
Finally, there are two couples (2)
Therefore, the probability we have to calculate is (120-96)/120=1/5
It means that the answer is A
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Re: Probability- Tough one ! [#permalink]
04 Aug 2005, 05:07
mbassmbass04 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 adjacents chairs?
(A) 1/5
(B) 2/5
(C) 4/5
(D) 5/7
(E) 3/4
Please see http://www.gmatclub.com/phpbb/viewtopic ... 9976#89976 and http://www.gmatclub.com/phpbb/viewtopic ... gle+person for some elegant explainations.
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Senior Manager
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ah, thanks for the posting Arsene_Wenger
I misread the problem as having two couplese being together.
(couple A) (couple B) single
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Senior Manager
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To solve this problem we need to find the total # of ways in which at least one couple is seated together and subtract it from the total# of outcomes.
Total # of outcomes = 5! = 120
If A = # of ways couple 1 is seated together and B = #of ways couple 2 is seated together, then we need N(A U B)
N (AUB) = N(A) + N(B) - N(A int B)
N(A) = N(B) = 2*4! = 48
N (A int B) = 2*2*3! = 24
N(AUB) = 48 + 48 - 24 = 72
Favorable outcomes = 120-72 = 48
P = 48/120 = 2/5
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