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:

A committee of three people is to be chosen from 4 married [#permalink]
18 Nov 2007, 19:15

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

69% (01:56) correct
31% (01:43) wrong based on 88 sessions

A committee of three people is to be chosen from 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16 B. 24 C. 26 D. 30 E. 32

Here's my solution so far :

Total combination 8C3 = 56

Choose the first couple 4C1 = 4 Choose the third person= 8C1 = 8 * 7! / 7! = 8

8 * 4 = 32 - Not sure if the answer should be 32 or 24. Please help.

Re: PS - Combination [#permalink]
20 Nov 2007, 12:50

alimad wrote:

A committee of three people is to be chosen from 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

16 24 26 30 32

Here's my solution so far :

Total combination 8C3 = 56

Choose the first couple 4C1 = 4 Choose the third person= 8C1 = 8 * 7! / 7! = 8

8 * 4 = 32 - Not sure if the answer should be 32 or 24. Please help.

following your solution:
Total combination 8C3 = 56
Choose the first couple 4C1 = 4
Choose the third person= 6C1 = 6
so # of comittees with a couple = 4x6 = 24
so # of comittees without a couple = 56 - 24 = 32

A committee of three people is to be chosen from 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

16 24 26 30 32

Soln: Now the first of the 3 people commitee can be chosen from any of the 8 people in 8 ways Now the second of the 3 people commitee can be chosen from any of the 6 people in 6 ways(leaving out the one person who is the couple of the person chosen in the first statement) Now the third of the 3 people commitee can be chosen from any of the 4 people in 4 ways(leaving out the two people whose couples have already been chosen)

Thus the total number of permutations is = 8 * 6 * 4 But since this is a selection, hence order does not matter

Therefore the total number of selections is = 8 * 6 * 4/3! = 32 ways

Sorry...I am not able to move this post to PS section. Could someone please move it to PS section.

Approach I thought is as follows...if some shorter method is possible please explain..

total selections = 8C3 = 56

let's say that couple is always present in this committee of three.

This means that there are 4 ways to select 2 people of the committee. ( 4 couples and any one couple can be selected in 4 ways) The third person can be selected out of remaining 6 people in 6 ways.

Therefore when couple exists there are: 4X6 = 24 ways

Re: PS-Combinations [#permalink]
11 May 2010, 14:24

Expert's post

A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee? A. 16 B. 24 C. 26 D. 30 E. 32

One of the approaches:

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 4 to send only one "representatives" to the committee: 4C3=4.

But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Re: PS-Combinations [#permalink]
17 Jun 2010, 01:16

Hi everybody, I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.

i can choose first member of the comitee in 8 ways.. removing the spouse of the selected person second member can be chosen in 6 ways.... third member in 4 ways..... so 8*6*4 which is not answer can someone explain why?

Re: PS-Combinations [#permalink]
17 Jun 2010, 04:50

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

amitjash wrote:

Hi everybody, I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.

i can choose first member of the comitee in 8 ways.. removing the spouse of the selected person second member can be chosen in 6 ways.... third member in 4 ways..... so 8*6*4 which is not answer can someone explain why?

The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people - 3! --> 192/3!=32.

Consider this: there are two couples and we want to choose 2 people not married to each other. Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible:

The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people - 3! --> 192/3!=32.

Thanks the two of you!

This is also the way I like to solve such questions and I believe it is way faster than any 10C3... and so on!

Re: A committee of three people is to be chosen from 4 married [#permalink]
12 Aug 2014, 03:54

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

There is one comment that stands out; one conversation having made a great impression on me in these first two weeks. My Field professor told a story about a...

Our Admissions Committee is busy reviewing Round 1 applications. We will begin sending out interview invitations in mid-October and continue until the week of November 9th, at which point...