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Two couples and one single person are seated at random in a [#permalink]
 20 Jan 2008, 11:21
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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 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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Re: couples in a row [#permalink]
20 Jan 2008, 11:47
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The answer is 2/5. I couldn't find a quick beautiful solution.
Let's numerate people 1,2,3,4,5 where (1,2) are a couple, (3,4) are a couple, 5 is a single person.
there is 1/5 probability that the single person is going to sit in chair one, then we need to put 1,2,3,4 in such a way that 2 people from the same couple don't sit next to each other. If we choose any person to sit in the second chair, his partner has to sit in chair 4 - probability of that is 1/3. Probability 1/5*1/3 = 1/15
there is 4/5 probability that a person from 1-4 is going to sit in chair one. Let's say it's person 1. then it's either person 5 who is going to sit in chair 2 (probability of that 1/4, then person 2 has to sit in chair 4 - probability of that 1/3 - total probability 4/5*1/4*1/3 = 1/15) or either 3 or 4 [say 3](probability of that 2/4 = 1/2, then his partner can't sit in chair 3 - probability of that 2/3 - total probability 4/5*1/2*2/3 = 4/15)
1/15+1/15+4/15 = 6/15=2/5. Not a beautiful solution but it took me less than 1.5 min to solve it.
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Re: couples in a row [#permalink]
20 Jan 2008, 12:07
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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 Answer 2/5
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Re: couples in a row [#permalink]
20 Jan 2008, 13:56
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Dp=1-\frac{P^4_4*P^2_2+P^4_4*P^2_2-P^3_3*P^2_2*P^2_2}{P^5_5}=1-\frac{4!*2+4!*2-3!*2*2}{5!}=1-\frac{16-4}{20}=\frac{2}{5}
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Re: couples in a row [#permalink]
20 Jan 2008, 15:32
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 where's this question from marco?
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Re: couples in a row [#permalink]
20 Jan 2008, 22:13
Thanks guys , so many different approaches for same question.
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Re: couples in a row [#permalink]
21 Jan 2008, 12:25
walker wrote: D
p=1-\frac{P^4_4*P^2_2+P^4_4*P^2_2-P^3_3*P^2_2*P^2_2}{P^5_5}=1-\frac{4!*2+4!*2-3!*2*2}{5!}=1-\frac{16-4}{20}=\frac{2}{5} walker, can you explain me this formula in details? thanks
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Re: couples in a row [#permalink]
21 Jan 2008, 13:36
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marcodonzelli wrote: walker wrote: D
p=1-\frac{P^4_4*P^2_2+P^4_4*P^2_2-P^3_3*P^2_2*P^2_2}{P^5_5}=1-\frac{4!*2+4!*2-3!*2*2}{5!}=1-\frac{16-4}{20}=\frac{2}{5} walker, can you explain me this formula in details? thanks - the number of permutations of the first couple and 3 single persons P^2_2 - the number of permutations of 2 single persons in the first couple P^4_4*P^2_2 - the total number of permutations in which the first couple sits together. P^4_4*P^2_2 - the total number of permutations in which the second couple sits together. P^3_3*P^2_2*P^2_2 - the total number of permutations in which the first couple sits together and the second couple sits together. P^4_4*P^2_2+P^4_4*P^2_2-P^3_3*P^2_2*P^2_2 - the total number of permutations in which the first couple sits together or the second couple sits together. P^5_5 - the total number of permutations of 5 single persons
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Re: couples in a row [#permalink]
24 Jan 2008, 09:01
wow this is pretty wild. 
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Re: couples in a row [#permalink]
26 Jan 2008, 19:43
This is not Manhattan, this is from Princeton Review.
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Re: couples in a row [#permalink]
27 Jan 2008, 10:56
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D
I also did by first finding the probability of couples being together and then subtracting from 1.
2 couples together and 1 single [A1A2][B1B2][C] = 3!*2*2 = 24 ways
3! for ABC and 2 each for interchanging positions between each couple
1 couple and 3 singles (subtract ways that have been counted in above) = [A1A2]BCD = [4! * 2] - 24 = 24 ways (this needs to be multiplied by 2 as we have two couples). So, 24*2=48
Total Number of ways = 5!
So, probability = 1-(48+24)/5! = 1-3/5=2/5
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Re: couples in a row [#permalink]
20 Mar 2008, 10:44
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GMAT TIGER wrote: walker :- math guru. a bit  I just return in quant to relax after hard (very hard!) work in Verbal....
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Re: couples in a row [#permalink]
20 Mar 2008, 10:46
walker wrote: GMAT TIGER wrote: walker :- math guru. a bit I just return in quant to relax after hard (very hard!) work in Verbal.... trade your for your math skills. 
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Re: couples in a row [#permalink]
20 Mar 2008, 10:58
bmwhype2 wrote: trade your for your math skills. agree 
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Re: couples in a row [#permalink]
24 Aug 2008, 14:01
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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 Total possible ways = 5! Two couple sits together [A1A2][B1B2]C = 2! * 2!*3! =24 only first couple sit together [A1A2] B1 B2 C = 2!*4! - 24 (substract 24 two couple sits together) =24 only second couple sit together A1A2 [B1 B2] C =2!*4! - 24 =24 probability = 1 - 24*3/120 = 1-3/5 =2/5
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Re: couples in a row [#permalink]
18 Feb 2009, 03:37
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probability = 1 - 24*3/120 = 1-3/5 =2/5
Total possible ways of selecting 5 persons in different ways is = 5!
Two couple sits together [A1A2][B1B2]C. So it forms only 3 groups. we can arrange the groups in 3! ways. we have 2 ways to arrange the person in a couple. so we have 2!=2
and so the calculation is Two couple sits together [A1A2][B1B2]C
= 2! * 2!*3! =24
only first couple sit together [A1A2] B1 B2 C. so it forms 4 groups (the couple and the other persons). we can arrange the different groups in 4! ways. we have 2 ways to arrange the person in a couple. so we have 2!=2
= 2!*4! - 24 (substract 24 two couple sits together)
=24
only second couple sit together A1A2 [B1 B2] C, so it forms 4 groups. we can arrange the different groups in 4! ways.
=2!*4! - 24 =24
probability = 1 - 24*3/120 = 1-3/5 =2/5
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Re: couples in a row [#permalink]
26 Sep 2009, 22:54
x2suresh 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 Total possible ways = 5! Two couple sits together [A1A2][B1B2]C = 2! * 2!*3! =24 only first couple sit together [A1A2] B1 B2 C = 2!*4! - 24 (substract 24 two couple sits together) =24 only second couple sit together A1A2 [B1 B2] C =2!*4! - 24 =24 probability = 1 - 24*3/120 = 1-3/5 =2/5 can you please explain why are you substracing 24 in the above marked equation? Thanks
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Re: couples in a row [#permalink]
27 Sep 2009, 02:57
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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
Soln:
Lets take the positions to be _ _ _ _ _
Now considering that the single person sits in seat 1.
The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Considering single person in seat 2
The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Considering single person in seat 3
The remaining seats can be filled in = 4 * 2 * 2 * 1 = 16 ways
Considering single person in seat 4
The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Considering single person in seat 5
The remaining seats can be filled in = 4 * 2 * 1 * 1 = 8 ways
Total number of ways in which 5 people can be arranged is 5! ways
thus we have probability
= (8 * 4 + 16) /5!
= 2/5
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Re: couples in a row [#permalink]
27 Sep 2009, 23:30
prabu wrote: x2suresh 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 Total possible ways = 5! Two couple sits together [A1A2][B1B2]C = 2! * 2!*3! =24 only first couple sit together [A1A2] B1 B2 C = 2!*4! - 24 (substract 24 two couple sits together) =24 only second couple sit together A1A2 [B1 B2] C =2!*4! - 24 =24 probability = 1 - 24*3/120 = 1-3/5 =2/5 can you please explain why are you substracing 24 in the above marked equation? Thanks I also kinda need the same answer.....
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Re: couples in a row [#permalink]
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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Re: couples in a row
[#permalink]
27 Sep 2009, 23:48
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