Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 20 Apr 2015, 22:58

### 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

# If a committee of 3 people is to be selected from among 5

Author Message
TAGS:
Intern
Joined: 04 Aug 2010
Posts: 15
Followers: 0

Kudos [?]: 15 [0], given: 0

If a committee of 3 people is to be selected from among 5 [#permalink]  04 Aug 2010, 03:06
6
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

72% (01:44) correct 28% (01:16) wrong based on 310 sessions
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Feb 2012, 13:35, edited 1 time in total.
Edited the question
Math Expert
Joined: 02 Sep 2009
Posts: 27000
Followers: 4169

Kudos [?]: 40035 [7] , given: 5397

Re: Combination/Permutation problem, couples [#permalink]  04 Aug 2010, 03:20
7
KUDOS
Expert's post
3
This post was
BOOKMARKED
kwhitejr wrote:
Can anyone demonstrate the following?

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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) each to send one "representative" to the committee: 5C3=10.

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

Total # of ways: 5C3*2^3=80.

Similar problems:
a-committee-of-three-people-is-to-be-chosen-from-four-teams-130617.html
if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html
a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
if-a-committee-of-3-people-is-to-be-selected-from-among-88772.html
a-comittee-of-three-people-is-to-be-chosen-from-four-married-130475.html
a-committee-of-three-people-is-to-be-chosen-from-4-married-101784.html
a-group-of-10-people-consists-of-3-married-couples-and-113785.html
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html

Hope it's clear.
_________________
Intern
Joined: 04 Aug 2010
Posts: 15
Followers: 0

Kudos [?]: 15 [0], given: 0

Re: Combination/Permutation problem, couples [#permalink]  04 Aug 2010, 03:26
Looking at the couples at first as single units was the eye-opener. Thanks very kindly.
Intern
Joined: 24 Feb 2012
Posts: 33
Followers: 0

Kudos [?]: 8 [3] , given: 18

new [#permalink]  29 Feb 2012, 10:01
3
KUDOS
5C3 --> ways 3 couples can be selected out of 5 couples = 10
(2C1)^3 --> ways to select 1 adult from each couple, for each of the three couples. = 8
Total Possibilities = 5C3 * (2C1)^3 = 80
Manager
Joined: 22 Feb 2012
Posts: 93
Schools: HBS '16
GMAT 1: 740 Q49 V42
GMAT 2: 670 Q42 V40
GPA: 3.47
WE: Corporate Finance (Aerospace and Defense)
Followers: 2

Kudos [?]: 20 [0], given: 25

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  29 Feb 2012, 10:16
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5420
Location: Pune, India
Followers: 1322

Kudos [?]: 6678 [2] , given: 176

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  29 Feb 2012, 10:24
2
KUDOS
Expert's post
AbeinOhio wrote:
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?

You have to select a committee of 3 people. It means A, B, C form the same committee as B, C, A i.e. you don't have to arrange people. But when you say 'first person' can be selected in 10 ways, you are arranging the 3 people in first, second and third spots. So according to you, A, B, C and B, C, A are different committees.
So all you need to do in un-arrange. To arrange 3 people, you multiply by 3!
To un-arrange, you will divide by 3!
480/3! = 80

Check out this post for a detailed explanation:
http://www.veritasprep.com/blog/2011/11 ... nstraints/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Last edited by VeritasPrepKarishma on 29 Feb 2012, 10:26, edited 1 time in total.
Math Expert
Joined: 02 Sep 2009
Posts: 27000
Followers: 4169

Kudos [?]: 40035 [1] , given: 5397

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  29 Feb 2012, 10:25
1
KUDOS
Expert's post
AbeinOhio wrote:
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?

You've done everything right, just missed the last step: since the people in the group needn't be ordered (arranged) in it, then you should divide 10*8*6=480 by 3! to get rid of duplication and get un-ordered groups --> 480/3!=80.

Hope it's clear.
_________________
Current Student
Joined: 24 Feb 2012
Posts: 6
Location: United States
Concentration: Finance, General Management
Schools: BYU (A)
GMAT Date: 03-07-2012
GPA: 3.5
WE: Law (Law)
Followers: 0

Kudos [?]: 5 [0], given: 1

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  29 Feb 2012, 16:25
Bunuel wrote:
AbeinOhio wrote:
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?

You've done everything right, just missed the last step: since the people in the group needn't be ordered (arranged) in it, then you should divide 10*9*8=480 by 3! to get rid of duplication and get un-ordered groups --> 480/3!=80.

Hope it's clear.

I think you mean 10 * 8 * 6 = 480.

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

Kudos [?]: 7 [0], given: 118

Re: Combination/Permutation problem, couples [#permalink]  05 Nov 2012, 21:51
Bunuel wrote:
kwhitejr wrote:
Can anyone demonstrate the following?

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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) to send only one "representatives" to the committee: 5C3=10.

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

Total # of ways: 5C3*2^3=80.

Similar problems:
ps-combinations-94068.html
ps-combinations-101784.html
committee-of-88772.html
if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html

Hope it's clear.

How it could be two person, out of three couples they have to sent 3 rep rite.. I am totally confused kindly enlighten me
Math Expert
Joined: 02 Sep 2009
Posts: 27000
Followers: 4169

Kudos [?]: 40035 [0], given: 5397

Re: Combination/Permutation problem, couples [#permalink]  06 Nov 2012, 02:43
Expert's post
breakit wrote:
Bunuel wrote:
kwhitejr wrote:
Can anyone demonstrate the following?

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

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) to send only one "representatives" to the committee: 5C3=10.

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

Total # of ways: 5C3*2^3=80.

Similar problems:
ps-combinations-94068.html
ps-combinations-101784.html
committee-of-88772.html
if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html

Hope it's clear.

How it could be two person, out of three couples they have to sent 3 rep rite.. I am totally confused kindly enlighten me

Yes, 3 couples should send total of 3 representatives, but EACH couple has 2 options (either husband or wife), thus these 3 couples can send 3 persons in 2^3=8 ways.

Hope it's clear.
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 17

Kudos [?]: 253 [1] , given: 11

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  27 Dec 2012, 18:27
1
KUDOS
kwhitejr wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120

I always use the ANAGRAM technique.

How many ways can we select three persons from 5 couples? Since we can only get one representative from each couple selected, we can imagine the 5 couples as 5 persons.

5!/3!2! = 10

Now, we know that there are 10 ways to select three persons of representing couples. But whenever we make selections, there are two ways to select from the couple, either wife or husband. Thus, 2 x 2 x 2 = 8

8*10 = 80

_________________

Impossible is nothing to God.

Manager
Joined: 13 Oct 2012
Posts: 78
Schools: IE '15 (A)
GMAT 1: 760 Q49 V46
Followers: 1

Kudos [?]: -14 [1] , given: 0

If a committee of 3 people is to be selected from among 5 marrie [#permalink]  07 Jan 2013, 10:43
1
KUDOS
5C3 - select three couples
2*2*2 --> select one member from each couple

ans - 5C3 * 8 = 80
Manager
Joined: 18 Oct 2011
Posts: 91
Location: United States
Concentration: Entrepreneurship, Marketing
GMAT Date: 01-30-2013
GPA: 3.3
Followers: 2

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

Re: GMATprep PS1 [#permalink]  07 Jan 2013, 10:52
Find the number of ways you can have a couple in the committee and then subtract from the total number of 3-person committees.
10C3 = 120 possible combinations
Since there are 5 couples and the final spot could be filled with any of the 8 remaining people you have 8x5 = 40 ways to achieve this.
Therefore you have 120-40 = 80 committees w/out a married couple. Answer: D
Manager
Joined: 02 Apr 2012
Posts: 72
Location: Argentina
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Followers: 1

Kudos [?]: 29 [0], given: 154

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  15 Jul 2013, 08:11
1
This post was
BOOKMARKED
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

(1) Combinatorial approach:

$$C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80$$

(2) Reversal combinatorial approach:

Total number of groups: $$C^10_3 = 120$$
Total number of groups with married people: $$C^5_1 * C^4_1 * C^2_1 = 5*4*2 = 40$$

120 - 40 = 80
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Senior Manager
Joined: 17 Dec 2012
Posts: 395
Location: India
Followers: 19

Kudos [?]: 241 [2] , given: 10

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  15 Jul 2013, 17:44
2
KUDOS
1
This post was
BOOKMARKED
Maxirosario2012 wrote:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

(1) Combinatorial approach:

$$C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80$$

i.e., IF AB, CD, EF, GH, IJ are the couples you have to select only one from a group

1. So selecting one from a group of 2 can be done in $$2C1$$ways. i.e., in 2 ways You have to select 3 people that way. So the total number of possibilities is $$2*2*2 = 8$$

2. Each group of 2 itself has to be selected from 5 such groups. You are selecting 3 groups of 2 from 5 such groups. therefore the total number of possibilities for this is $$5C3= 10$$.

3. $$(1) * (2) = 8* 10 = 80.$$
Quote:
(2) Reversal combinatorial approach:

Total number of groups: $$C^10_3 = 120$$
Total number of groups with married people: $$C^5_1 * C^4_1 * C^2_1 = 5*4*2 = 40$$

120 - 40 = 80

1. Total number of possibilities of selecting 3 people out of 10 people$$= 10C3= 120$$
2. For total number of groups with a married couple the situation is you have (i)2 people who are married to each other i.e., a group of 2 married to each other and (ii) one other person. The number of possibilities of (i) is the number of ways one group of married people can be selected from 5 groups of married people which is $$5C1 = 5$$ways, the number of possibilities of (ii) can be arrived by finding out in how many ways the remaining person can be selected. It can be done in $$8C1=8$$ ways because if you remove the selected married couple 8 persons will remain.
3. Total number of ways of having a married couple$$in the group of 3 = 5*8=40$$4. So number of groups in which 2 people are not married couple $$= 120-40=80.$$
_________________

Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravna.com

Classroom Courses in Chennai
Free Online Material

Manager
Joined: 12 Feb 2012
Posts: 109
Followers: 1

Kudos [?]: 14 [0], given: 28

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  21 Jul 2013, 18:38
AbeinOhio wrote:
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?

What if you started out by choosing the first person in 1 one way instead of 10.

So you pick a person of the 10, doesn't matter who. For the next slot you have 8 choices. and the next 6. so (1)(8)(6). Why isn't this approach valid?
Math Expert
Joined: 02 Sep 2009
Posts: 27000
Followers: 4169

Kudos [?]: 40035 [0], given: 5397

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  21 Jul 2013, 21:05
Expert's post
alphabeta1234 wrote:
AbeinOhio wrote:
I'm pretty lost on this one..

Here was the logic that i thought would work but obviously I am missing something:

5 couples = 10 people

So first position there are 10 options

Second position, the first person's mate is excluded now so instead of having 9 options you have 8 options

3rd position the first and 2nd person's mates are both excluded now so instead of having 7 options you have 6 options

so I got 10 * 9 * 8 = 480

What am I missing?

What if you started out by choosing the first person in 1 one way instead of 10.

So you pick a person of the 10, doesn't matter who. For the next slot you have 8 choices. and the next 6. so (1)(8)(6). Why isn't this approach valid?

Why do you have 1 choice and not 10?
Why do you have 8 choices for the second pick and not 9?
Why do you have 6 choices for the third pick and not 8?

The reason why AbeinOhio's solution is not correct is explained here: if-a-committee-of-3-people-is-to-be-selected-from-among-98533.html#p1051830

Hope it helps.
_________________
Intern
Joined: 20 Aug 2013
Posts: 2
Followers: 0

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

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  21 Aug 2013, 16:40

5C1*2C1* 4C1*2C1* 8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5420
Location: Pune, India
Followers: 1322

Kudos [?]: 6678 [0], given: 176

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  21 Aug 2013, 21:58
Expert's post
brunawang wrote:

5C1*2C1* 4C1*2C1* 8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot

Ok, here is what is wrong with your solution.
Say, the couples are (A1, A2), (B1, B2), (C1, C2), (D1, D2) and (E1, E2)

Now you cannot have 2 people from the same couple.

Two different scenarios in your solution:

You select one couple in 5 ways. Say you selected (C1, C2). In two ways you selected one of them. You got C2.
You select one couple in 4 ways now. Say you selected ((E1, E2). In two ways you selected one of them. You got E2.
You selected one person out of 8 in 8 ways, You got A2.

You select one couple in 5 ways. Say you selected (E1, E2). In two ways you selected one of them. You got E2.
You select one couple in 4 ways now. Say you selected ((C1, C2). In two ways you selected one of them. You got C2.
You selected one person out of 8 in 8 ways, You got A2.

Notice that they give you the same team but you have counted these two as different selections. Hence your answer is incorrect.
When making a selection, try to use 5C3 method. It helps you think clearly. You select 3 couples out of the 5. Now from each couple you select one person out of the two. So you get 5C3*2*2*2 - there is no double counting here.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Veritas Prep GMAT course is coming to India. Enroll in our weeklong Immersion Course that starts March 29!

Veritas Prep Reviews

Intern
Joined: 20 Aug 2013
Posts: 2
Followers: 0

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

Re: If a committee of 3 people is to be selected from among 5 [#permalink]  22 Aug 2013, 04:01
VeritasPrepKarishma wrote:
brunawang wrote:

5C1*2C1* 4C1*2C1* 8C1=640, being:

5C1*2C1 - ways of choosing one couple out of 5, multiplied by 2C1 ways of choosing one person of a couple
4C1*2C1 - ways of choosing another couple out of 4, multiplied by 2C1 ways of choosing one person of a couple
8C1 - people that can occupy the third spot

Ok, here is what is wrong with your solution.
Say, the couples are (A1, A2), (B1, B2), (C1, C2), (D1, D2) and (E1, E2)

Now you cannot have 2 people from the same couple.

Two different scenarios in your solution:

You select one couple in 5 ways. Say you selected (C1, C2). In two ways you selected one of them. You got C2.
You select one couple in 4 ways now. Say you selected ((E1, E2). In two ways you selected one of them. You got E2.
You selected one person out of 8 in 8 ways, You got A2.

You select one couple in 5 ways. Say you selected (E1, E2). In two ways you selected one of them. You got E2.
You select one couple in 4 ways now. Say you selected ((C1, C2). In two ways you selected one of them. You got C2.
You selected one person out of 8 in 8 ways, You got A2.

Notice that they give you the same team but you have counted these two as different selections. Hence your answer is incorrect.
When making a selection, try to use 5C3 method. It helps you think clearly. You select 3 couples out of the 5. Now from each couple you select one person out of the two. So you get 5C3*2*2*2 - there is no double counting here.

Thanks Karishma, your explanation was perfect!
Re: If a committee of 3 people is to be selected from among 5   [#permalink] 22 Aug 2013, 04:01

Go to page    1   2    Next  [ 26 posts ]

Similar topics Replies Last post
Similar
Topics:
If a committee of 3 people is to be selected from among 5 6 13 Oct 2007, 07:15
If a committee of 3 people is to be selected from among 5 1 12 Aug 2007, 11:53
If a committee of 3 people is to be selected from among 5 5 22 Jul 2006, 15:44
If a committee of 3 people is to be selected from among 5 11 01 Dec 2005, 13:33
If a committee of 3 people is to be selected from among 5 3 21 May 2005, 19:43
Display posts from previous: Sort by