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:

6 married couples are present at a party. if 4 people are [#permalink]

Show Tags

06 Aug 2006, 15:05

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

6 married couples are present at a party. if 4 people are selected out of these 12, what is the probability that none of these people will be married to each other?

A)1/33

B) 2/33

C) 1/3

D) 16/33

E) 11/12

Please explain your answer.
_________________

for every person who doesn't try because he is
afraid of loosing , there is another person who
keeps making mistakes and succeeds..

4 can be selected out of 12 in 12C4=12*11*10*9/1*2*3*4=55*9=495 ways

If A is selected as one of 4
Then 3 can be selected from remaining 11 in 11C3 =11*10*9/1*2*3=11*5*3=165 ways of which likelyhood of B being in selection is =165/11 = 15
Therefore 150/165= 10/11 is the probability that if A is first, B is not in the list of 4.
not sure how to proceed...