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:
If a committee of 3 people is to be selected from among 5 [#permalink]
04 Aug 2010, 03:06
2
This post received KUDOS
9
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
25% (medium)
Question Stats:
73% (01:44) correct
27% (01:20) wrong based on 411 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?
Re: Combination/Permutation problem, couples [#permalink]
04 Aug 2010, 03:20
7
This post received KUDOS
Expert's post
7
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.
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
Re: If a committee of 3 people is to be selected from among 5 [#permalink]
29 Feb 2012, 10:24
3
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
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
Re: If a committee of 3 people is to be selected from among 5 [#permalink]
29 Feb 2012, 10:25
2
This post received 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.
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.
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.
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.
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.
Re: If a committee of 3 people is to be selected from among 5 [#permalink]
27 Dec 2012, 18:27
1
This post received 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
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
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?
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
Re: If a committee of 3 people is to be selected from among 5 [#permalink]
15 Jul 2013, 17:44
2
This post received KUDOS
2
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?
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 \(= 120-40=80.\) _________________
Re: If a committee of 3 people is to be selected from among 5 [#permalink]
21 Aug 2013, 16:40
Can someone please help me? I don't know what I am doing wrong.
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
Re: If a committee of 3 people is to be selected from among 5 [#permalink]
21 Aug 2013, 21:58
Expert's post
1
This post was BOOKMARKED
brunawang wrote:
Can someone please help me? I don't know what I am doing wrong.
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. Your team (A2, C2, E2)
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. Your team (A2, C2, E2)
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. _________________
Re: If a committee of 3 people is to be selected from among 5 [#permalink]
22 Aug 2013, 04:01
VeritasPrepKarishma wrote:
brunawang wrote:
Can someone please help me? I don't know what I am doing wrong.
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. Your team (A2, C2, E2)
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. Your team (A2, C2, E2)
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!
gmatclubot
Re: If a committee of 3 people is to be selected from among 5
[#permalink]
22 Aug 2013, 04:01
You know what’s worse than getting a ding at one of your dreams schools . Yes its getting that horrid wait-listed email . This limbo is frustrating as hell . Somewhere...
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Wow! MBA life is hectic indeed. Time flies by. It is hard to keep track of the time. Last week was high intense training Yeah, Finance, Accounting, Marketing, Economics...