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

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04 Aug 2010, 04:06

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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?

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

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

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]

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29 Feb 2012, 17:25

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.

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.

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]

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27 Dec 2012, 19:27

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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

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]

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15 Jul 2013, 09:11

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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?

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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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]

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21 Aug 2013, 17:40

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

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]

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22 Aug 2013, 05:01

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.

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