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by bblast » Mon Jun 20, 2011 8:13 am
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?


1/5
1/4
3/8
2/5
1/2


my solution : total ways 120.
both coupls always together C1c1 S K1k1 = 3!*2!*2!

thus prob that couple always together = 24/120

therefore prob that never together = 1-1/5 = 4/5

please point where I erred

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by Frankenstein » Mon Jun 20, 2011 8:18 am
Hi,
You are removing only the case of both couples together. So, you are essentially counting the case of one couple sitting together and none of the couples siting together. So, you need to remove the case of 'one of the couples sitting together' as well.
Last edited by Frankenstein on Mon Jun 20, 2011 8:31 am, edited 1 time in total.
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by Frankenstein » Mon Jun 20, 2011 8:29 am
If you want to solve this by using complement method, here is the way:
Let them be a1,a2,b1,b2,c
Case-1: a1,a2 are together(This includes the case of b1,b2 together as well)
No. of ways is (4!).(2!)=48
Case-2: b1,b2 are together(This includes the case of b1,b2 together as well)
No. of ways is 4!.2!=48
You have already calculated the number of ways in which both couples are together as 3!.2!.2! = 24 ways
So, the number of ways in which atleast one couple sits together is 48+48-24(because both together is counted twice) = 72
So, probability that neither of them are together is 1-(72/120) = 2/5

Hence, D
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by Surev » Mon Jun 20, 2011 8:50 am
Let the people be denoted as CA1, CA2, S, CB1, CB2 where ca1 and ca2 is the first couple and cb1 and cb2 is the second couple and s is the singleton.
Now s can take all 5 position and the rest will have to adjust accordingly to avoid their other half. Therefore the possibilities are:

S _ _ _ _ = 1*4*2*1*1 = 8 (1st is the single, second can be any one of the four, third can be any one of the other two and fourth and fifth will be fixed for the remaining two)
_ S _ _ _ = 4*1*2*1*1 = 8
_ _ S _ _ = 4*2*1*2*1 = 16
_ _ _ S _ = 8 (Similar to 2nd above)
_ _ _ _ S = 8 (Similar to 1st above)

So total favorable results = 8 + 8 + 16 + 8 + 8 = 48
Total possible outcomes = 5! = 120

Probability = 48/120 = 2/5
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by pemdas » Mon Jun 20, 2011 10:29 am
first clue one couple (A1,A2) as A and consider A B1 B2 C, or C(4,1)=24
then clue another couple (B1,B2) -> C(4,1)=24, hence there (24+24) or 48 ways two couples sit together out of possible 5! ways

we can say that the probability two couples don't sit together in adjacent places is 48/5!=
48/120=2/5
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by GMATGuruNY » Mon Jun 20, 2011 1:59 pm
bblast 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?


1/5
1/4
3/8
2/5
1/2


my solution : total ways 120.
both coupls always together C1c1 S K1k1 = 3!*2!*2!

thus prob that couple always together = 24/120

therefore prob that never together = 1-1/5 = 4/5

please point where I erred

oa- [spoiler]2/5 :oops: [/spoiler]
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