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:

4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

18 Oct 2010, 11:40

1

This post received KUDOS

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

41% (01:42) correct 59% (02:16) wrong based on 272 sessions

HideShow timer Statistics

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?

a. 3/24 b. 9/24 c. 15/24 d. 18/24 e. 21/24

(I am getting a different result than the OA - a result that even does not appear in the answer choices. Please collaborate)

Ways all 4 sit in the correct seat = 1 Ways such that exactly 3 in correct seat = 0 (If there is one person in the wrong seat, then whose seat it is must also be in the wrong seat) Ways such that exactly 2 in wrong seat = C(4,2) = 6 (choose the two in wrong seat, only one way to get it wrong) Ways such that exactly 3 in wrong seat = C(4,1) x (3! - C(3,2) - 1) = 4x2= 8 (choose the one in the correct seat x the ways 3 guys can be in 3 wrong seats -- which is again all ways to seat 3 people - ways in which all correct (1) - ways in which 2 wrong (C(3,2)) )

So total ways to get all 4 wrong = 4! - 1 - 0 - 6 - 8 = 9

So probability = 9/24 Answer : (b)

This is a tough question, and the lieklihood of getting it on the GMAT is very low
_________________

Man - you played it! Thanks. I didn't think of all the possibilities.

My initial solution was:

3/4 probability for the first person to NOT sit on his chair 2/3 probability for the second person to NOT sit on his chair (from the remaining 3 chairs) 1/2 probability for the third person to NOT sit on his chair (from the remaining 2 chairs)

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?

a. 3/24 b. 9/24 c. 15/24 d. 18/24 e. 21/24

(I am getting a different result than the OA - a result that even does not appear in the answer choices. Please collaborate)

Man - you played it! Thanks. I didn't think of all the possibilities.

My initial solution was:

3/4 probability for the first person to NOT sit on his chair 2/3 probability for the second person to NOT sit on his chair (from the remaining 3 chairs) 1/2 probability for the third person to NOT sit on his chair (from the remaining 2 chairs)

So, my final solution was: 3/4 * 2/3 * 1/2 = 1/4

But I missed some arrangements.

Thanks

+1 from me

hi....I also tried this method but later realized that we are making the probability of the last one to choose wrong seat as 1 in this method. (3/4 * 2/3 * 1/2 *1) But that is not true.

The first 3 guys can interchange their seats, all sit in wrong seats and the last one can still sit in his own place. We are missing something here.

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

25 Jul 2013, 14:47

You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

08 Oct 2013, 22:47

R0dman wrote:

You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

I am getting (9+6)/24 as answer?

Which is equal to 15/24 = 5/8

Bunuel, Can you Please comment? _________________

Like my post Send me a Kudos It is a Good manner. My Debrief: http://gmatclub.com/forum/how-to-score-750-and-750-i-moved-from-710-to-189016.html

You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

09 Oct 2013, 05:44

honchos wrote:

R0dman wrote:

You can also solve it as 1 - ((1/4)+(3/4*1/3)+(3/4*1/3*1/2)). This gives the exact same answer. The part that's in brackets is the probability of negative outcome (Someone sitting on their chair);

I am getting (9+6)/24 as answer?

Which is equal to 15/24 = 5/8

Bunuel, Can you Please comment?

Once again. it's "1- negative outcome". Probability of negative outcome is in brackets. So it goes (1/4+1/4+1/8) = (6/24+6/24+3/24) =15/24. Then you deduct it from 1. So basically you getting 1- 15/24 = 9/24.

I dont know whether this is the right way to approach the problem , however i get the same ans.

Originally seated A B C D

now when after they get up and when they sit back again . 1st- A has option to sit on 3 seats ( apart from his previous seat . thus he now sits on B's seat.) 2nd- Similarly B has option to sit on 3 seats ( because A has already occupied B's previous seat, thus B sits on a's seat.) 3rd- Now C has only 1 option to sit on D's seat . and similarly D also has one option to sit on C's seat.)

hence total favourable outcomes 3*3*1*1=9

and total possible outcomes =4!=24

probability of the favourable outcome= 9/24.

Please correct me if i am wrong.

eladshus wrote:

Man - you played it! Thanks. I didn't think of all the possibilities.

My initial solution was:

3/4 probability for the first person to NOT sit on his chair 2/3 probability for the second person to NOT sit on his chair (from the remaining 3 chairs) 1/2 probability for the third person to NOT sit on his chair (from the remaining 2 chairs)

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

28 Oct 2013, 17:17

eladshus wrote:

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?

A. 3/24 B. 9/24 C. 15/24 D. 18/24 E. 21/24

There is a formula for total number of possible de arrangements:

Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own envelope is = n! [ (1/2!) - (1/3!) + (1/4!) ...... (-1)^n (1/n!) ]

In this case n=4. Therefore, total number of possible dearrangements such that none of them sit in their place = 4! [ (1/2!) - (1/3!) + (1/4!) ] = 9.

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

31 Oct 2013, 06:17

Hi guys, this is my first post. Here is my way to the solution: when the 4 people come back the first one has 3/4 possibilities to choose a seat different from the previous one, let's call him A and assume he seats were B used to seat before; then B for sure will be sitting on a different place, because A is now on his place, so B's possibility to choose a different seat is 1; C will have 2 choices, and just one suitable for sitting on a different seat, so C's possibility will be 1/2; D will have no choices, because C is now on his seat, so also D will have possibility 1 to sit on a different seat. 3/4 * 1 * 1/2 * 1 = 3/8 that is the same of 9/24

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

31 Oct 2013, 09:02

bugatti wrote:

Hi guys, this is my first post. Here is my way to the solution: when the 4 people come back the first one has 3/4 possibilities to choose a seat different from the previous one, let's call him A and assume he seats were B used to seat before; then B for sure will be sitting on a different place, because A is now on his place, so B's possibility to choose a different seat is 1; C will have 2 choices, and just one suitable for sitting on a different seat, so C's possibility will be 1/2; D will have no choices, because C is now on his seat, so also D will have possibility 1 to sit on a different seat. 3/4 * 1 * 1/2 * 1 = 3/8 that is the same of 9/24

Is there any flaw in the above?

Thanks a lot

let's call him A and assume he seats were B used to seat before;, what if A doent sit where B sat?

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

31 Oct 2013, 16:28

Quote:

what if A doent sit where B sat?

Yes you are right, that's the flaw. By changing the choice made by A at the beginning changes the probability as a whole. For instance if A sits where C used to seat, then B will have 2/3 of possibilities to sit on a different seat. It doesn't work

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

17 Nov 2013, 08:39

I got the correct answer, its simple, this problem is based on the Derangement Formula. Search the net and read about it, it comes to use in many problem scenarios.
_________________

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

02 Jun 2014, 08:20

saurabhprashar wrote:

eladshus wrote:

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?

A. 3/24 B. 9/24 C. 15/24 D. 18/24 E. 21/24

There is a formula for total number of possible de arrangements:

Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own envelope is = n! [ (1/2!) - (1/3!) + (1/4!) ...... (-1)^n (1/n!) ]

In this case n=4. Therefore, total number of possible dearrangements such that none of them sit in their place = 4! [ (1/2!) - (1/3!) + (1/4!) ] = 9.

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?

A. 3/24 B. 9/24 C. 15/24 D. 18/24 E. 21/24

There is a formula for total number of possible de arrangements:

Number of ways in which n letters can be placed in n directed letters so that no letter goes into its own envelope is = n! [ (1/2!) - (1/3!) + (1/4!) ...... (-1)^n (1/n!) ]

In this case n=4. Therefore, total number of possible dearrangements such that none of them sit in their place = 4! [ (1/2!) - (1/3!) + (1/4!) ] = 9.

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

02 Jun 2014, 15:59

killer question..

X - Y - Z - V <--- Original Arrangement (below are the potential seats each person can sit in to satisfy constraint) * - X - X - X Y - * - Y - Y Z - Z - * - Z V - V - V - *

starting from the first non-underlined X, you can pick any combo of the below letters a long as they don't occupy the same column (i.e. seat). there are 3 such arrangements for each X or 9 total (divided by 4! or 24 total possibilities)
_________________

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

07 Nov 2015, 10:07

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.
_________________

Re: 4 people are sitting in a 4 seat row watching a football [#permalink]

Show Tags

20 Jul 2017, 06:33

eladshus wrote:

4 people are sitting in a 4 seat row watching a football game. At halftime they all get up. When they return, they each randomly sit down on one of the 4 chairs. What is the likelihood that none of the 4 end up sitting in the same chair that they sat in during the first half?