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:

A committee of 3 people is to be chosen from four married [#permalink]
11 May 2010, 11:35

4

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

68% (01:47) correct
32% (01:13) wrong based on 424 sessions

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?

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

9

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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

7

This post received KUDOS

Expert's post

3

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:

A_1,B_1; A_1,B_2; A_2,B_1; A_2,B_2.

Only 4 such committees are possible.

If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

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: permutation problem [#permalink]
28 Oct 2010, 21:45

5

This post received KUDOS

1

This post was BOOKMARKED

So there are 3 people we need to choose from 8.

Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways

Re: permutation problem [#permalink]
29 Oct 2010, 01:28

5

This post received KUDOS

Since there are 4 couples so we have 8 people involved.

The First person can be selected from the 8 people in 8 ways The second person should not be a spouse of the first and hence we have 6 ways to choose him/her The Third person should not be a spouse of either of the 2, so we can choose him in 6 ways.

So the total no. of ways we can choose the people will be 8*6*4 ways. However since order is not important (i.e A,B,C is the same as B,A,C) so we divide the total ways by 3!

Hence the total number of groups is 32 _________________

Please give me kudos, if you like the above post. Thanks.

Re: permutation problem [#permalink]
09 Apr 2012, 13:27

devashish wrote:

Since there are 4 couples so we have 8 people involved.

The First person can be selected from the 8 people in 8 ways The second person should not be a spouse of the first and hence we have 6 ways to choose him/her The Third person should not be a spouse of either of the 2, so we can choose him in 6 ways.

So the total no. of ways we can choose the people will be 8*6*4 ways. However since order is not important (i.e A,B,C is the same as B,A,C) so we divide the total ways by 3!

Hence the total number of groups is 32

Hey,

can you help me how you get the 3!. What does it stand for or what does this number say?

Re: permutation problem [#permalink]
10 Apr 2012, 03:31

Expert's post

1

This post was BOOKMARKED

andih wrote:

devashish wrote:

Since there are 4 couples so we have 8 people involved.

The First person can be selected from the 8 people in 8 ways The second person should not be a spouse of the first and hence we have 6 ways to choose him/her The Third person should not be a spouse of either of the 2, so we can choose him in 6 ways.

So the total no. of ways we can choose the people will be 8*6*4 ways. However since order is not important (i.e A,B,C is the same as B,A,C) so we divide the total ways by 3!

Hence the total number of groups is 32

Hey,

can you help me how you get the 3!. What does it stand for or what does this number say?

Thanks

It seems that you need to brush up your fundamentals:

Re: A committee of 3 people is to be chosen from four married [#permalink]
06 Aug 2012, 10:21

Person (p1 p2 p3 p4 p5 p6 p7 p8) No of ways to choose 1st Person: Any 8 No of ways to choose 2nd Person: 6 (Pair of 1st person can not be considered so we need to exclude 1 pair) No of ways to choose 3rd Person: 4 (Pair of 1st & 2nd Person can not be considered so we need to exclude 2 pair) No of ways : 8X6X4 (Now we have done a permutation) But here order of the team member is not important and 3 person can arrange themselves in 3! ways. So need to divide the permutation by 3!. Ans: 8*6*4/3! = 32 _________________

Re: permutation problem [#permalink]
05 Nov 2012, 22:28

shrouded1 wrote:

So there are 3 people we need to choose from 8.

Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways

Total ways = 4+4+12+12 = 32

Posted from my mobile device

not sure whether this is an easy way.. but i understood this quite well except the part marked in red.

is that like if 2 men has been already selected from 2 couples we are looking for women in two ways from other two couples.

Re: permutation problem [#permalink]
06 Nov 2012, 02:47

Expert's post

breakit wrote:

shrouded1 wrote:

So there are 3 people we need to choose from 8.

Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways

Total ways = 4+4+12+12 = 32

Posted from my mobile device

not sure whether this is an easy way.. but i understood this quite well except the part marked in red.

is that like if 2 men has been already selected from 2 couples we are looking for women in two ways from other two couples.

Exactly. If we choose 2 men out of 4, then the third person must be a woman from the remaining two couples: 4C2*2=6*2=12.

Re: permutation problem [#permalink]
06 Nov 2012, 06:23

Bunuel wrote:

breakit wrote:

shrouded1 wrote:

So there are 3 people we need to choose from 8.

Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways

Total ways = 4+4+12+12 = 32

Posted from my mobile device

not sure whether this is an easy way.. but i understood this quite well except the part marked in red.

is that like if 2 men has been already selected from 2 couples we are looking for women in two ways from other two couples.

Exactly. If we choose 2 men out of 4, then the third person must be a woman from the remaining two couples: 4C2*2=6*2=12.

Hope it's clear.

4C2*2=6*2=12. ??? (4*3*2*1)/(2*1) = 12 but you have mentioned as 6.. is something I am missing here.

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

Almost half of MBA is finally coming to an end. I still have the intensive Capstone remaining which started this week, but things have been ok so far...