It is currently 22 Nov 2017, 04:34

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

A committee of 3 people is to be chosen from four married

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

7 KUDOS received
Director
Director
User avatar
Joined: 03 Sep 2006
Posts: 864

Kudos [?]: 1109 [7], given: 33

GMAT ToolKit User
A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 11 May 2010, 12:35
7
This post received
KUDOS
46
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

66% (00:42) correct 34% (01:15) wrong based on 1357 sessions

HideShow timer Statistics

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

[Reveal] Spoiler:
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

Thus no couple = 8C3 - (4X6) = 32
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Oct 2013, 05:44, edited 2 times in total.
Edited the question and added the OA

Kudos [?]: 1109 [7], given: 33

Expert Post
11 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133019 [11], given: 12402

Re: PS-Combinations [#permalink]

Show Tags

New post 11 May 2010, 15:24
11
This post received
KUDOS
Expert's post
31
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.

Total # of ways: 4C3*2^3=32.

Answer: E.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 133019 [11], given: 12402

1 KUDOS received
Manager
Manager
avatar
Joined: 17 Mar 2010
Posts: 171

Kudos [?]: 211 [1], given: 9

Re: PS-Combinations [#permalink]

Show Tags

New post 17 Jun 2010, 02:16
1
This post received
KUDOS
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?

Kudos [?]: 211 [1], given: 9

Expert Post
21 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133019 [21], given: 12402

Re: PS-Combinations [#permalink]

Show Tags

New post 17 Jun 2010, 05:50
21
This post received
KUDOS
Expert's post
7
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.

You can see the similar problem at:
committee-of-88772.html#p669797

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 133019 [21], given: 12402

Manager
Manager
avatar
Joined: 22 Jun 2010
Posts: 56

Kudos [?]: 78 [0], given: 10

Re: PS-Combinations [#permalink]

Show Tags

New post 22 Jul 2010, 08:37
Bunuel wrote:

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!

Kudos [?]: 78 [0], given: 10

11 KUDOS received
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 793

Kudos [?]: 1211 [11], given: 25

Location: London
GMAT ToolKit User Reviews Badge
Re: permutation problem [#permalink]

Show Tags

New post 28 Oct 2010, 22:45
11
This post received
KUDOS
5
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

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

Posted from my mobile device
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 1211 [11], given: 25

17 KUDOS received
Manager
Manager
avatar
Joined: 16 Jun 2010
Posts: 180

Kudos [?]: 103 [17], given: 5

Re: permutation problem [#permalink]

Show Tags

New post 29 Oct 2010, 02:28
17
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.

Kudos [?]: 103 [17], given: 5

TOEFL Forum Moderator
avatar
Joined: 16 Nov 2010
Posts: 1603

Kudos [?]: 601 [0], given: 40

Location: United States (IN)
Concentration: Strategy, Technology
Premium Member Reviews Badge
Re: PS-Combinations [#permalink]

Show Tags

New post 25 Apr 2011, 04:58
4C3 = 4 (# of ways couples can be chosen)

=> total # of ways = 4 * 2 * 2 * 2 (2 options from each couple)

= 32

Answer - E
_________________

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

GMAT Club Premium Membership - big benefits and savings

Kudos [?]: 601 [0], given: 40

Intern
Intern
avatar
Joined: 14 Feb 2012
Posts: 40

Kudos [?]: 43 [0], given: 13

Location: Germany
Concentration: Technology, Strategy
GMAT Date: 06-13-2012
Re: permutation problem [#permalink]

Show Tags

New post 09 Apr 2012, 14:27
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?

Thanks

Kudos [?]: 43 [0], given: 13

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133019 [0], given: 12402

Re: permutation problem [#permalink]

Show Tags

New post 10 Apr 2012, 04:31
Expert's post
2
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:

Factorial: http://mathworld.wolfram.com/Factorial.html
Combinatorics: math-combinatorics-87345.html

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 133019 [0], given: 12402

3 KUDOS received
Senior Manager
Senior Manager
avatar
Joined: 15 Jun 2010
Posts: 357

Kudos [?]: 461 [3], given: 50

Schools: IE'14, ISB'14, Kellogg'15
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)
Reviews Badge
Re: A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 06 Aug 2012, 11:21
3
This post received
KUDOS
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
_________________

Regards
SD
-----------------------------
Press Kudos if you like my post.
Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

Kudos [?]: 461 [3], given: 50

Manager
Manager
avatar
Joined: 28 Sep 2011
Posts: 67

Kudos [?]: 33 [0], given: 10

Location: United States
GMAT 1: 520 Q34 V27
GMAT 3: 690 Q47 V38
GPA: 3.01
WE: Information Technology (Commercial Banking)
GMAT ToolKit User
Re: A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 06 Aug 2012, 14:14
The method I used to avoid combinations

Line up the people: 12 34 56 78

Then we have something like this:
135
136
137
138
145
146
147
148

So for the "1's" we have 8 numbers. For "2" we will also have 8, and another 8 more for both "3" and "4".

8 * 4 = 32

Kudos [?]: 33 [0], given: 10

Intern
Intern
avatar
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 48

Kudos [?]: 9 [0], given: 118

Concentration: General Management, Leadership
GMAT Date: 08-27-2013
GPA: 3.8
Re: permutation problem [#permalink]

Show Tags

New post 05 Nov 2012, 23:28
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.

Kudos [?]: 9 [0], given: 118

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133019 [0], given: 12402

Re: permutation problem [#permalink]

Show Tags

New post 06 Nov 2012, 03:47
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 133019 [0], given: 12402

Intern
Intern
avatar
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 48

Kudos [?]: 9 [0], given: 118

Concentration: General Management, Leadership
GMAT Date: 08-27-2013
GPA: 3.8
Re: permutation problem [#permalink]

Show Tags

New post 06 Nov 2012, 07:23
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.

Kudos [?]: 9 [0], given: 118

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133019 [0], given: 12402

Re: permutation problem [#permalink]

Show Tags

New post 06 Nov 2012, 07:29

Kudos [?]: 133019 [0], given: 12402

Intern
Intern
avatar
Status: wants to beat the gmat
Joined: 18 Jul 2012
Posts: 20

Kudos [?]: 9 [0], given: 1

Location: United States
Re: A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 06 Nov 2012, 12:53
is this a correct way to get the answer? or was it just coincidence:

8C3 - 4(4C2) = 56 - 4(6) = 32

Kudos [?]: 9 [0], given: 1

Expert Post
1 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133019 [1], given: 12402

Re: A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 07 Nov 2012, 05:37
1
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
watwazdaquestion wrote:
is this a correct way to get the answer? or was it just coincidence:

8C3 - 4(4C2) = 56 - 4(6) = 32


It's not clear what is the logic behind the formula.

Reversed approach would be:
There are 8C3=56 ways to select 3 people out of 8 without any restriction;
There are 4C1*6=24 ways there to be a couple among 3 members: 4C1 ways to select a couple out of 4, which will be in the committee and 6 ways to select the third remaining member (since there will be 6 members left after we select a couple out of 8 people).

56-24=32.

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 133019 [1], given: 12402

Senior Manager
Senior Manager
User avatar
Joined: 13 Aug 2012
Posts: 458

Kudos [?]: 558 [0], given: 11

Concentration: Marketing, Finance
GPA: 3.23
GMAT ToolKit User
Re: A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 27 Dec 2012, 21:54
LM wrote:
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



How many ways to select 3 represented couples from 4 couples? 4!/3!1! = 4
How many ways to select a person from a pair? 2
\(=4 * 2 * 2 * 2 = 32\)

Answer: E

More detailed explanation here : Selection/Deselection Technique
_________________

Impossible is nothing to God.

Kudos [?]: 558 [0], given: 11

Current Student
avatar
Joined: 18 Jun 2013
Posts: 6

Kudos [?]: 7 [0], given: 5

Location: United States
Concentration: Marketing, Strategy
GMAT 1: 540 Q39 V27
GMAT 2: 640 Q44 V35
GPA: 3.59
Re: A committee of 3 people is to be chosen from four married [#permalink]

Show Tags

New post 02 Sep 2013, 23:55
I solved it using this method, hope I'm using the correct concept

total 4 couples = 8 people in total
total no of ways to choose 3 people out of 8 = 8!/(5!3!) = 56
No. of ways couples are included in the com = 4! = 24
Therefore no. of couples with no couples included = 56-24 = 32.

Kudos [?]: 7 [0], given: 5

Re: A committee of 3 people is to be chosen from four married   [#permalink] 02 Sep 2013, 23:55

Go to page    1   2    Next  [ 39 posts ] 

Display posts from previous: Sort by

A committee of 3 people is to be chosen from four married

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.