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 350,000 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:

There are 5 married couples and a group of three is to be [#permalink]
06 Feb 2005, 16:55

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)

I can't seem to get 80. Help appreciated. Thank you.

You need to use total combination minus the special cases where a couple was picked.

So you pick a couple. Which is C(5,1). Now you still need one more person to form the 3 people committee from the remaing 8 people. There are C(8,1) ways of doing this. In other words the number of combinations where a couple was picked is C(5,1)*C(8,1).

You need to use total combination minus the special cases where a couple was picked.

So you pick a couple. Which is C(5,1). Now you still need one more person to form the 3 people committee from the remaing 8 people. There are C(8,1) ways of doing this. In other words the number of combinations where a couple was picked is C(5,1)*C(8,1).

...I was wondering if you could do it this way and would like some input:

AB CD EF GH IJ

represent 5 married couples

- - - <--- represents three spots

If you pick A you have 8 choices between the rest excluding B. This apllies to the other ten members. Therefore the answer is 8*10 = 80 Does this work? _________________

Since you need a committee of 3 select 3 couples from the 5, 5 choose 3 = 10 ways. Then there are 2 choices from each couple. Therefore the total arrangements is 10 * (2*2*2) = 80.

Re: Probability question [#permalink]
21 Feb 2005, 14:15

sk wrote:

There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group? (Ans. 80)

I can't seem to get 80. Help appreciated. Thank you.

HongHu's method makes all the sense, however I keep getting stuck with my approach:
10*8*6 = 480

ten ways to pick first person in a trio, 8 ways to pick one out of the remaining 8 people, and 6 ways to pick the last, third person.

lastochka,
After a long thought,I guess I figured the problem.

Let's say there is no restriction and wife&husband can be in the same group.Then the problem is down to picking 3 from 10. According to your logic, you will pick 10*9*8. However,it is obiously wrong, order is not important.M1,M2,M3 are same as M1,M3,M2.So it can not be permutations,it should be combinations.

Looking from the fundamental diff from permutations,in combinations, order is not important,so you devide the P by r! and you will get C.
nCr = nPr/r!

Similarly in the original example,since you picked the perumations way, you need to devide it by r! which is 3! .So 10*8*6/3!=80.

Thanks for raising the question thus helping refresh my rusty math concepts.

lastochka, After a long thought,I guess I figured the problem.

Let's say there is no restriction and wife&husband can be in the same group.Then the problem is down to picking 3 from 10. According to your logic, you will pick 10*9*8. However,it is obiously wrong, order is not important.M1,M2,M3 are same as M1,M3,M2.So it can not be permutations,it should be combinations.

Looking from the fundamental diff from permutations,in combinations, order is not important,so you devide the P by r! and you will get C. nCr = nPr/r!

Similarly in the original example,since you picked the perumations way, you need to devide it by r! which is 3! .So 10*8*6/3!=80.

Thanks for raising the question thus helping refresh my rusty math concepts.

Since you need a committee of 3 select 3 couples from the 5, 5 choose 3 = 10 ways. Then there are 2 choices from each couple. Therefore the total arrangements is 10 * (2*2*2) = 80.

Yes, a good way to solve this kind of question.

700Plus wrote:

Looking from the fundamental diff from permutations,in combinations, order is not important,so you devide the P by r! and you will get C. nCr = nPr/r!

Similarly in the original example,since you picked the perumations way, you need to devide it by r! which is 3! .So 10*8*6/3!=80.

Another great approach!

A lot of times there are multiple approaches to a permutation/combination question. If you have time it may be good that you could use different methods to verify your solution.

How the growth of emerging markets will strain global finance : Emerging economies need access to capital (i.e., finance) in order to fund the projects necessary for...

One question I get a lot from prospective students is what to do in the summer before the MBA program. Like a lot of folks from non traditional backgrounds...