Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Oct 2016, 07:13

# Today:

Live Q&A: UCLA Adcom at 8AM PDT in Chat1 | LBS/INSEAD at 9AM in Chat1 | Expecting Interview Invites from Ross: Join Chat2 for live updates

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 9

Kudos [?]: 459 [3] , given: 0

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

### Show Tags

20 Jan 2008, 11:21
3
KUDOS
21
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

37% (03:04) correct 63% (02:14) wrong based on 366 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA
Director
Joined: 01 Jan 2008
Posts: 629
Followers: 5

Kudos [?]: 170 [1] , given: 1

Re: couples in a row [#permalink]

### Show Tags

20 Jan 2008, 11:47
1
KUDOS
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.
Director
Joined: 12 Jul 2007
Posts: 862
Followers: 15

Kudos [?]: 274 [8] , given: 0

Re: couples in a row [#permalink]

### Show Tags

20 Jan 2008, 12:07
8
KUDOS
1
This post was
BOOKMARKED
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

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 519

Kudos [?]: 3326 [2] , given: 360

Re: couples in a row [#permalink]

### Show Tags

20 Jan 2008, 13:56
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
D

$$p=1-\frac{P^4_4*P^2_2+P^4_4*P^2_2-P^3_3*P^2_2*P^2_2}{P^5_5}=1-\frac{4!*2+4!*2-3!*2*2}{5!}=1-\frac{16-4}{20}=\frac{2}{5}$$
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Joined: 15 Nov 2007
Posts: 135
Followers: 1

Kudos [?]: 48 [0], given: 2

Re: couples in a row [#permalink]

### Show Tags

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?
Manager
Joined: 20 Sep 2007
Posts: 106
Followers: 1

Kudos [?]: 66 [0], given: 0

Re: couples in a row [#permalink]

### Show Tags

20 Jan 2008, 22:13
Thanks guys , so many different approaches for same question.
VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 9

Kudos [?]: 459 [0], given: 0

Re: couples in a row [#permalink]

### Show Tags

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
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 519

Kudos [?]: 3326 [4] , given: 360

Re: couples in a row [#permalink]

### Show Tags

21 Jan 2008, 13:36
4
KUDOS
Expert's post
2
This post was
BOOKMARKED
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

$$P^4_4$$ - 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
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

CEO
Joined: 21 Jan 2007
Posts: 2756
Location: New York City
Followers: 10

Kudos [?]: 806 [0], given: 4

Re: couples in a row [#permalink]

### Show Tags

24 Jan 2008, 09:01
wow this is pretty wild.
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Director
Joined: 01 May 2007
Posts: 792
Followers: 1

Kudos [?]: 270 [0], given: 0

Re: couples in a row [#permalink]

### Show Tags

26 Jan 2008, 19:43
This is not Manhattan, this is from Princeton Review.
Manager
Joined: 02 Jan 2008
Posts: 159
Followers: 2

Kudos [?]: 107 [5] , given: 0

Re: couples in a row [#permalink]

### Show Tags

27 Jan 2008, 10:56
5
KUDOS
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
CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 519

Kudos [?]: 3326 [1] , given: 360

Re: couples in a row [#permalink]

### Show Tags

20 Mar 2008, 10:44
1
KUDOS
Expert's post
GMAT TIGER wrote:
walker :- math guru.

a bit
I just return in quant to relax after hard (very hard!) work in Verbal....
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

CEO
Joined: 21 Jan 2007
Posts: 2756
Location: New York City
Followers: 10

Kudos [?]: 806 [0], given: 4

Re: couples in a row [#permalink]

### Show Tags

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....

_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

CEO
Joined: 17 Nov 2007
Posts: 3589
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 519

Kudos [?]: 3326 [0], given: 360

Re: couples in a row [#permalink]

### Show Tags

20 Mar 2008, 10:58
bmwhype2 wrote:

agree
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

SVP
Joined: 07 Nov 2007
Posts: 1820
Location: New York
Followers: 32

Kudos [?]: 810 [10] , given: 5

Re: couples in a row [#permalink]

### Show Tags

24 Aug 2008, 14:01
10
KUDOS
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
_________________

Smiling wins more friends than frowning

Intern
Joined: 31 Jul 2008
Posts: 1
Followers: 0

Kudos [?]: 2 [1] , given: 0

Re: couples in a row [#permalink]

### Show Tags

18 Feb 2009, 03:37
1
KUDOS
1
This post was
BOOKMARKED
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
Manager
Joined: 09 Aug 2009
Posts: 53
Followers: 1

Kudos [?]: 6 [0], given: 1

Re: couples in a row [#permalink]

### Show Tags

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
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 136 [3] , given: 3

Re: couples in a row [#permalink]

### Show Tags

27 Sep 2009, 02:57
3
KUDOS
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
Manager
Joined: 05 Jul 2009
Posts: 182
Followers: 1

Kudos [?]: 46 [0], given: 5

Re: couples in a row [#permalink]

### Show Tags

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.....
Manager
Joined: 09 Aug 2009
Posts: 53
Followers: 1

Kudos [?]: 6 [0], given: 1

Re: couples in a row [#permalink]

### Show Tags

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
Re: couples in a row   [#permalink] 27 Sep 2009, 23:48

Go to page    1   2    Next  [ 26 posts ]

Similar topics Replies Last post
Similar
Topics:
9 3 persons (1 couple and 1 single) are seated at random in a 7 20 Apr 2012, 16:59
26 Two couples and one single person are seated at random in a 8 10 Jan 2012, 07:01
71 Two couples and one single person are seated at random in a 24 21 Nov 2009, 13:36
6 Two couples and one single person are seated at random in a 12 25 Sep 2009, 14:07
8 Two couples and one single person are seated at random in a 9 30 Nov 2007, 08:52
Display posts from previous: Sort by