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

 It is currently 29 Jun 2016, 13:44

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

Probability

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Senior Manager
Joined: 01 May 2004
Posts: 336
Location: USA
Followers: 1

Kudos [?]: 75 [0], given: 0

Show Tags

14 May 2004, 18:20
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hello!
Guys, try to solve this problem

9 people, including 3 couples are to be seated in a row of 9 chairs. What is the probability that

a. None of the Couples are together
b. Only one couple is together
c. All the couples are together
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 12 [0], given: 0

Show Tags

15 May 2004, 01:02
boksana wrote:
Hello!
Guys, try to solve this problem

9 people, including 3 couples are to be seated in a row of 9 chairs. What is the probability that

a. None of the Couples are together
b. Only one couple is together
c. All the couples are together

Hi, boksana!

All combinations: 9!

Combinations for which all the couples are together:

6!*2*2*2 (because there are 6! ways to arrange 6 entities - 3 couples and 3 other persons, and 2 ways of sitting a couple MW and WM).

=> p(c) = 6!*8/9! = 1/63.

Combinations for which 2 couples are together and 1 is not:

7!*2*2 - 6!*2*2*2 (the same logic - 7 entities can be seated in 7! ways, and there are 2 ways in each couple).

=> p(.) = (7!*4-6!*8)/9! = 1/18-1/63=5/126.

Combinations for which only 1 couple is:

8!*2 - [previous 2 cases together] = 8!*2 - 7!*4.

p(b) = [8!*2-7!*4]/9! = 2/9-1/18 = 1/6.

Combinations for which there are no couples:

p(a) = 1 - p(b) - p(.) - p(c) = 1 - 1/6 - 5/126 - 1/63 = 98/126 = 7/9.
Senior Manager
Joined: 01 May 2004
Posts: 336
Location: USA
Followers: 1

Kudos [?]: 75 [0], given: 0

Show Tags

15 May 2004, 11:03
Good job!
It's all correct.
Manager
Joined: 07 May 2004
Posts: 183
Location: Ukraine, Russia(part-time)
Followers: 2

Kudos [?]: 12 [0], given: 0

Show Tags

15 May 2004, 11:26
boksana wrote:
Good job!
It's all correct.

I know...

Are there any questions? Proposals? Suggestions?

Feel free to ask if you have any.
Hi!   [#permalink] 15 May 2004, 11:26
Display posts from previous: Sort by

Probability

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.