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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

20 Jan 2008, 10:21

2

This post received KUDOS

24

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

37% (03:05) correct
63% (02:18) wrong based on 389 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?

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.

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.

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?

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
_________________

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

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
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

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

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

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

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 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

gmatclubot

Re: couples in a row
[#permalink]
27 Sep 2009, 22:48

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...