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

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If a committee of 3 people is to be selected from among 5 [#permalink] New post 04 Aug 2010, 03:06
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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.
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Re: Combination/Permutation problem, couples [#permalink] New post 04 Aug 2010, 03:20
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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.

Answer: D.

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

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new [#permalink] New post 29 Feb 2012, 10:01
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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
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 15 Jul 2013, 17:44
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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 2C1ways. 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.
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(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 = 5ways, 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 couplein the group of 3 = 5*8=404. So number of groups in which 2 people are not married couple = 120-40=80.

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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 29 Feb 2012, 10:24
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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

There's your answer.

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

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Last edited by VeritasPrepKarishma on 29 Feb 2012, 10:26, edited 1 time in total.
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 27 Dec 2012, 18:27
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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

Answer: C

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

ans - 5C3 * 8 = 80
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Re: Combination/Permutation problem, couples [#permalink] New post 04 Aug 2010, 03:26
Looking at the couples at first as single units was the eye-opener. Thanks very kindly.
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 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?
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 29 Feb 2012, 10:25
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.

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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 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!
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Re: Combination/Permutation problem, couples [#permalink] New post 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.

Answer: D.

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
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Re: Combination/Permutation problem, couples [#permalink] New post 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.

Answer: D.

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.

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Re: GMATprep PS1 [#permalink] New post 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
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 15 Jul 2013, 08:11
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

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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 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?
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 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.

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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 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
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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 21 Aug 2013, 21:58
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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.

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Re: If a committee of 3 people is to be selected from among 5 [#permalink] New post 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! :)
Re: If a committee of 3 people is to be selected from among 5   [#permalink] 22 Aug 2013, 04:01
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