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:

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

Please provide your time in solving this problem along with your answer:

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?

I think I have seen this type of question before.
Say 8 persons are A,B,C,D,E,F,G,H.. AB,CD,EF,GH are couples
Calculate by reducing number of ways when couples will be together from number of total ways.
Number of total ways..8C3
Number of ways when couples will be together.. Say A, B are together..in this case number of ways to select 3rd person from remaining 6 is 6C1=6
Similarly when C,D are together..number of ways to selct 3rd person is 6.
When E, F are together..another 6 ways.
G,H are together..another 6 ways..
So total 24 ways.. when couples are together.
Hence required number of ways = 8C3 -24
=56-24
= 32
So 32 should be the answer

Consider the couples as (H1,W1), (H2,W2), (H3,W3), (H4,W4), (H5,W5)
For each pair of couple they can pair up with either the husband or wife of another pair to form a group of 3
So for example, H1,W1,H2, H1,W1,W2..etc. So one couple can have total of 8 possible pairings
5 couples will give 5*8=40 pairs

So total number of 3 people group with husband and wife not in the saem group = 120-40 = 80

10 members. So, 10C3 = 120 is the number of groups formed. This 120 may contain even the couples.

Groups that contain couples only = 5C1 * 8C1 = 40

So, the groups that donot contain couples are 120-40 = 80

lastochka wrote:

Please provide your time in solving this problem along with your answer:

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?

1) Total - unwanted.
10C3 - 5*8 = 80
5 is for choosing one married couple of the 5, and 8 is for choosing another person out of the remaining8 .

2) 10*8*6/3! = 80.
For the first person you have 10 options.
For the second, only 8.
For the third, only 6.
Divide by 3! because the order doesn't matter.

mallelac, do you mind explaining how you came to the formula for the undesireable outcomes? Wouldn't 5C1+8C1 only give you the number of people that would comprise a 2 person team?

The undesirable outcome should have a couple. We can choose one out of 5 couples in 5C1 ways thus filling two spots(not one spot). Now, 10-2 = 8 members can be fill up the third spot in 8C1 ways. Thus total ways of undesired outcomes = 5C1*8C1 = 40.

Do let me know if I am not clear.

abennett wrote:

mallelac, do you mind explaining how you came to the formula for the undesireable outcomes? Wouldn't 5C1+8C1 only give you the number of people that would comprise a 2 person team?