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:

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other. There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - \(C^6_4*(C^2_1)^4\)

look at other problem:

Quote:

Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team?

we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

total ways of choosing 4 out of 12:12x11x10x9 first person can be chosen in 12 ways. The next person can be chosen in 10 ways (because we don't want spouses to be in the group). The next person can be again chosen in only 8 ways (out of the 10 people left, we have to exclude 2 whose spouses we have already selected). And the last person in 6 ways.

suresh, can you please explain the logic of this calculation? Thank you!

12C1 = select any person from 6 married couples (12 person) 10C1 = select second person from remaining people and exclude the first person's spouse 8C1 = select 3rd person from remain people exclue first and second perssons's spouses 6C1 = select 4th person from remaining people exclude 1st,2nd ,3rd persons's spouse

Becuase order is not matter.. you need to divide by 4!

Did you get it?

=
_________________

Your attitude determines your altitude Smiling wins more friends than frowning

Given that there are 6 married couples. If we select only 4 people out of the 12, what is the probability that none of them are married to each other?

So, we should count how many possibilities we have to form group of 4 people with restriction: none of them are married to each other. There a few ways to count all combinations. One of the ways is following: 4 people belong to 4 distinct couples. Therefore, we could choose these 4 couples and then 1 person out of each couple - \(C^6_4*(C^2_1)^4\)

look at other problem:

Quote:

Given that there are 8 soccer teams. If we select only 6 people out of the 88 (8 teams, 11 people in each team), what is the probability that none of them are out of the same team?

we can use the same reasoning: choose 6 teams out of 8 teams (our 6 people are from 6 different teams) and then choose 1 player out of 11 for each team.

I hope it is clearer now.

Wow, makes very clear sense now. Much obliged. Ali

another way of looking at this problem :

total number of ways = 12C4 = 495

now let's try to find out total number of unfavourable ways. We can subtract those from 495 to get the number of favourable ways.

number of ways of choosing 4 people such that there are two couples = 6C2 = 15

number of ways of choosing 4 people such that there is only one couple = number of ways of choosing one couple * number of ways of choosing two people who are not couples = 6C1 (10C2 -5) = 240.

total number of unfavourable ways = 240 + 15 = 255

In x2suresh's solution, I'm unable to visualize the exact arrangements because of which we should divide 12x10x8x6 ways by 4!. Could someone please explain in more detail ? The thing is, we're choosing here, so how does the question of arrangements come in picture at all ?

Regards, Subhash
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

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...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...