Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 23 May 2017, 21:55

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Director
Joined: 03 Sep 2006
Posts: 875
Followers: 7

Kudos [?]: 891 [3] , given: 33

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

### Show Tags

11 May 2010, 12:35
3
KUDOS
38
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

66% (01:43) correct 34% (01:17) wrong based on 1075 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
Math Expert
Joined: 02 Sep 2009
Posts: 38844
Followers: 7719

Kudos [?]: 105942 [12] , given: 11601

### Show Tags

11 May 2010, 15:24
12
KUDOS
Expert's post
24
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.

_________________
Manager
Joined: 17 Mar 2010
Posts: 183
Followers: 3

Kudos [?]: 185 [1] , given: 9

### Show Tags

17 Jun 2010, 02:16
1
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?
Math Expert
Joined: 02 Sep 2009
Posts: 38844
Followers: 7719

Kudos [?]: 105942 [19] , given: 11601

### Show Tags

17 Jun 2010, 05:50
19
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.
_________________
Manager
Joined: 22 Jun 2010
Posts: 57
Followers: 1

Kudos [?]: 57 [0], given: 10

### Show Tags

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!
Retired Moderator
Joined: 02 Sep 2010
Posts: 803
Location: London
Followers: 110

Kudos [?]: 1020 [8] , given: 25

### Show Tags

28 Oct 2010, 22:45
8
KUDOS
4
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
_________________
Manager
Joined: 16 Jun 2010
Posts: 185
Followers: 2

Kudos [?]: 92 [11] , given: 5

### Show Tags

29 Oct 2010, 02:28
11
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.

Math Expert
Joined: 02 Sep 2009
Posts: 38844
Followers: 7719

Kudos [?]: 105942 [0], given: 11601

### Show Tags

16 Dec 2010, 15:56
Merging similar topics.

Also check this: confuseddd-99055.html
_________________
SVP
Joined: 16 Nov 2010
Posts: 1666
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 34

Kudos [?]: 533 [0], given: 36

### Show Tags

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

_________________

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

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 14 Feb 2012
Posts: 40
Location: Germany
Concentration: Technology, Strategy
GMAT Date: 06-13-2012
Followers: 0

Kudos [?]: 38 [0], given: 13

### Show Tags

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
Math Expert
Joined: 02 Sep 2009
Posts: 38844
Followers: 7719

Kudos [?]: 105942 [0], given: 11601

### Show Tags

10 Apr 2012, 04:31
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:

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

Hope it helps.
_________________
Senior Manager
Joined: 15 Jun 2010
Posts: 361
Schools: IE'14, ISB'14, Kellogg'15
WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)
Followers: 11

Kudos [?]: 398 [3] , given: 50

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

### Show Tags

06 Aug 2012, 11:21
3
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

Manager
Joined: 28 Sep 2011
Posts: 70
Location: United States
GMAT 1: 520 Q34 V27
GMAT 3: 690 Q47 V38
GPA: 3.01
WE: Information Technology (Commercial Banking)
Followers: 1

Kudos [?]: 31 [0], given: 10

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

### Show Tags

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
Manager
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

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

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 38844
Followers: 7719

Kudos [?]: 105942 [0], given: 11601

### Show Tags

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.
_________________
Manager
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

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

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 38844
Followers: 7719

Kudos [?]: 105942 [0], given: 11601

### Show Tags

06 Nov 2012, 07:29
breakit wrote:
4C2*2=6*2=12. ??? (4*3*2*1)/(2*1) = 12 but you have mentioned as 6.. is something I am missing here.

$$C^2_4=\frac{4!}{2!2!}=6$$.
_________________
Manager
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

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

### Show Tags

06 Nov 2012, 12:01
Bunuel wrote:
breakit wrote:
4C2*2=6*2=12. ??? (4*3*2*1)/(2*1) = 12 but you have mentioned as 6.. is something I am missing here.

$$C^2_4=\frac{4!}{2!2!}=6$$.

$$C^2_4=\frac{4!}{2!2!}=6$$.[/quote]

sorry , my mind is totally screwed with above stuff.. help me here
Math Expert
Joined: 02 Sep 2009
Posts: 38844
Followers: 7719

Kudos [?]: 105942 [0], given: 11601

### Show Tags

06 Nov 2012, 12:23
breakit wrote:
Bunuel wrote:
breakit wrote:
4C2*2=6*2=12. ??? (4*3*2*1)/(2*1) = 12 but you have mentioned as 6.. is something I am missing here.

$$C^2_4=\frac{4!}{2!2!}=6$$.

$$C^2_4=\frac{4!}{2!2!}=6$$.

sorry , my mind is totally screwed with above stuff.. help me here[/quote]

Hope it helps.
_________________
Intern
Status: wants to beat the gmat
Joined: 18 Jul 2012
Posts: 20
Location: United States
Followers: 0

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

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

### Show Tags

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
Re: A committee of 3 people is to be chosen from four married   [#permalink] 06 Nov 2012, 12:53

Go to page    1   2   3    Next  [ 47 posts ]

Similar topics Replies Last post
Similar
Topics:
24 A committee of three people is to be chosen from four teams 20 10 Mar 2016, 04:47
A comittee of three people is to be chosen from four married 1 09 Apr 2012, 14:08
18 A committee of three people is to be chosen from 4 married 19 21 Mar 2017, 09:30
7 A committee of three people is to be chosen from four 9 06 Aug 2012, 11:19
30 A committee of three people is to be chosen from four marrie 8 01 Oct 2013, 01:31
Display posts from previous: Sort by