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:

3 persons - one couple and one single are seated at random in a row of 5 seats

what is the probability that the couple does not sit together

please explain your approach

best regards

Let's find the probability that a couple sits together (right next to each other) and subtract that value from 1.

Total # of ways 3 persons \(C_1\), \(C_2\) and \(S\) to be seated in a row of 5 seats is \(\frac{5!}{2!}=60\). Consider this, we are interested in arrangement of \(C_1, \ C_2, \ S, \ E, \ E\), so in arrangement of 5 letters out of which 2 E's are identical (E denotes an empty seat);

# of ways for a couple to sit together is \(\frac{4!}{2!}*2=24\). Consider a couple as a single unit: \(\{C_1,C_2\}, \ S, \ E, \ E\), so we have total of 4 units out of which 2 E's are identical, # of arrangement of these units is \(\frac{4!}{2!}\), but \(C_1\), \(C_2\) within their unit can be arranged in 2 ways (\(\{C_1,C_2\}\) or \(\{C_2,C_1\}\)), so total # of arrangement for this case is \(\frac{4!}{2!}*2=24\);

Re: 3 persons (1 couple and 1 single) are seated at random in a [#permalink]

Show Tags

28 Dec 2012, 18:42

keiraria wrote:

3 persons (1 couple and 1 single) are seated at random in a row of 5 chairs. What is the probability that the couple does not sit together?

A. 5/7 B. 4/5 C. 2/5 D. 3/5 E. 11/18

Given: {H} {W} {S} will attempt to seat on _ _ _ _ _ seats

How many ways for {H} {W} {S} to seat on _ _ _ _ _ seats? 5*4*3 How many ways for {HW} {S} to seat together on _ _ _ _ _ seats? 4*3 Then multiply by 2 to account for the arrangement of HW. 4*3*2!

What is the probability of {H}{W} NOT seating together? \(=1 - \frac{4*3*2}{5*4*3} = 1 - \frac{2}{5} = 3/5\)

The single person is no different from an empty chair

Thus, there are \(\frac{{5*4}}{2}=10\) ways to pick two chairs for the couple, but only 4 in which they sit together (CCEEE, ECCEE, EECCE, EEECC).

1 - 4/10 = 3/5 is the answer.

The perspective we often use to solve such questions is this: The vacant seats are no different from two identical people. Assume that each vacant spot is taken by an imaginary person V.

'The single person is no different from an empty chair' is a refreshing perspective that we can use! Good point! Assume the rest of the three chairs are vacant. Since it is a probability question, the probability we will obtain will be correct. Mind you, the number of ways in which you can arrange a couple and an individual is not given by 10. It is given by 60 only (as shown by Bunuel above). But 10 is the number of ways in which we can choose 2 seats for a couple. In 4 of those 10 ways, the seats will be next to each other and in 6 cases they will not be. Hence the probability obtained will be 3/5. You can also think that you can make the husband and the wife sit in 2 of the 5 chairs in 5*4 = 20 ways. Out of these, in 4*2 = 8 ways, they will sit next to each other. In 12 ways, they will not sit next to each other. So probability will still remain 12/20 = 3/5

As discussed before, in probability questions, whatever logic you use to get the numerator, use the same logic to get the denominator.
_________________

Re: 3 persons (1 couple and 1 single) are seated at random in a [#permalink]

Show Tags

12 Oct 2015, 05:00

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

3 persons (1 couple and 1 single) are seated at random in a [#permalink]

Show Tags

23 Oct 2015, 21:15

why for this question are we doing 5!/2! for the chairs to people?

I recall from a circular question, when there was 4 men and 4 women sitting around the table, and we had to calculate the number of ways if all the women sat together.

which was (4-1)! * 4C1 * 4!

So why is this question 5!/2! instead of 5C3 ?

We can incorporate this question into my confusion surrounding this "http://gmatclub.com/forum/there-are-x-people-and-y-chairs-in-a-room-where-x-and-y-are-170525.html"

There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12. Since x and y are primes, then x=5 and y=7 OR x=7 and y=5.

If x=5 and y=7, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 chairs from 7, and 5! is the number of arrangements of 5 people on those chairs.

If x=7 and y=5, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 people who will get the chairs, and 5! is the number of arrangements of 5 people on those chairs.

why for this question are we doing 5!/2! for the chairs to people?

I recall from a circular question, when there was 4 men and 4 women sitting around the table, and we had to calculate the number of ways if all the women sat together.

which was (4-1)! * 4C1 * 4!

So why is this question 5!/2! instead of 5C3 ?

We can incorporate this question into my confusion surrounding this "http://gmatclub.com/forum/there-are-x-people-and-y-chairs-in-a-room-where-x-and-y-are-170525.html"

There are x people and y chairs in a room where x and y are positive prime numbers. How many ways can the x people be seated in the y chairs (assuming that each chair can seat exactly one person)?

(1) x + y = 12. Since x and y are primes, then x=5 and y=7 OR x=7 and y=5.

If x=5 and y=7, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 chairs from 7, and 5! is the number of arrangements of 5 people on those chairs.

If x=7 and y=5, the number of arrangements would be C57∗5!=7!2!, where C57 is the number of way to choose 5 people who will get the chairs, and 5! is the number of arrangements of 5 people on those chairs.

I think you are confusing between straight up arrangements of n things in a row = n! and arrangement of n people around a table = circular arrangement = (n-1)!

For the question : 3 people (1 couple and 1 single), we are doing 5!/2! to account for 5 'things' to arrange (C1,C2, S,E,E where Es are empty seats) and 2! accounts for the 2 empty seats.

You can not do 5C3 as this will remove the cases when you have EEC1C2S or EC1C2ES etc.

Re: 3 persons (1 couple and 1 single) are seated at random in a [#permalink]

Show Tags

28 Nov 2016, 19:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Since my last post, I’ve got the interview decisions for the other two business schools I applied to: Denied by Wharton and Invited to Interview with Stanford. It all...

Marketing is one of those functions, that if done successfully, requires a little bit of everything. In other words, it is highly cross-functional and requires a lot of different...