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 350,000 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]
12 Oct 2003, 17:08

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

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?

Answer : 2/5

i started from the basic formula

[n(A) + n(B)- n(A and B)+n(not A and not B) ] / (total # of ways) = 1

P(A) + P(B) - P(A and B) + P(not A and Not B) = 1

in our case
n(A) = # of ways one of the two couples seated together
n(B) = # of ways the second of the two couples can beseated together
n(A and B) = # of ways Both of the couples can be seated together
n(Not A and Not B) = # of ways None of the couples together

In all these cases, the total # of ways is 5!, so we divide the entire sum by 5!

We count the couple as a one single unit, as they have to be together.
so we now have a total of 4 "people". they can be arranged in 4! ways. now the couple itself can be arranged in 2! ways. The total number of possible arrangements is 5! ... it follows that

P(second couple seated together) => same as the above = 48/ 120

P(two couples seated together)
so now we have three "people". using the same logic as above,
3 people can be arranged in 3! ways. Note : Here we multiply first by 2 as the first couple can be arranged together in 2! ways...similarly the second couple can be seated together in 2! ways...and the total # of arrangements is 5!.
P(two couples seated together ) = 3! * 2 * 2 / 5 ! = 24 /120

So , it follows that
P(neither couple seated together) = 1 - 48/120 - 48/120 + 24/120 = 1- 72/120 = 48/120 = 2/5

anandnk you have calculated only the option when 2 couples are together and they can sit in 3! x2x2=24 ways. But we have to calculate also the option when 1 couple is together , then the possible ways for the couples to sit are 4!x2=48. So when 2 couples are together we have 24 ways and when 1 is together we have 48 or total 72 ways. Drawing venn diagrams is helpful here . the prob is 1-72/120=48/120=2/5

opps I got it.
But I have a question.
Let us say 1 couple C1 sits together then we have h2(husband2),w2(wife2), s(lonely person) members left
now we have C1, h1,w1,s people sitting in 4! ways and C1 can sit in 2 ways. so we have 2 * 4! ways

In this we also have the combination C1,h1,w1,s and C1,w1,h1,s. These
combinations account for 2 couples sitting together. Why do we again add 24( combinations in which 2 couples sit together) ?

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

I had an interesting conversation with a friend this morning, and I realized I need to add a last word on the series of posts on my application process. Five key words:...