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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]

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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?

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

Last edited by Bunuel on 01 Feb 2012, 14:35, edited 1 time in total.
Edited the question

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04 Aug 2010, 04:20
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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.

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.
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04 Aug 2010, 04:26
Looking at the couples at first as single units was the eye-opener. Thanks very kindly.

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29 Feb 2012, 11:01
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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]

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29 Feb 2012, 11:16
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]

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29 Feb 2012, 11:24
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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

Check out this post for a detailed explanation:
http://www.veritasprep.com/blog/2011/11 ... nstraints/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by VeritasPrepKarishma on 29 Feb 2012, 11:26, edited 1 time in total. Kudos [?]: 17324 [7], given: 232 Math Expert Joined: 02 Sep 2009 Posts: 41871 Kudos [?]: 128513 [3], given: 12179 Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 29 Feb 2012, 11:25 3 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. Hope it's clear. _________________ Kudos [?]: 128513 [3], given: 12179 Intern Joined: 24 Feb 2012 Posts: 6 Kudos [?]: 5 [0], given: 1 Location: United States Concentration: Finance, General Management Schools: BYU (A) GMAT Date: 03-07-2012 GPA: 3.5 WE: Law (Law) Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 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. Hope it's clear. I think you mean 10 * 8 * 6 = 480. Cheers! Kudos [?]: 5 [0], given: 1 Intern Status: K... M. G... Joined: 22 Oct 2012 Posts: 48 Kudos [?]: 9 [0], given: 118 Concentration: General Management, Leadership GMAT Date: 08-27-2013 GPA: 3.8 Re: Combination/Permutation problem, couples [#permalink] ### Show Tags 05 Nov 2012, 22:51 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 Kudos [?]: 9 [0], given: 118 Math Expert Joined: 02 Sep 2009 Posts: 41871 Kudos [?]: 128513 [0], given: 12179 Re: Combination/Permutation problem, couples [#permalink] ### Show Tags 06 Nov 2012, 03:43 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. _________________ Kudos [?]: 128513 [0], given: 12179 Senior Manager Joined: 13 Aug 2012 Posts: 459 Kudos [?]: 540 [2], given: 11 Concentration: Marketing, Finance GPA: 3.23 Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 27 Dec 2012, 19:27 2 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 8*10 = 80 Answer: C _________________ Impossible is nothing to God. Kudos [?]: 540 [2], given: 11 Manager Joined: 13 Oct 2012 Posts: 69 Kudos [?]: -9 [1], given: 0 Concentration: General Management, Leadership Schools: IE '15 (A) GMAT 1: 760 Q49 V46 If a committee of 3 people is to be selected from among 5 marrie [#permalink] ### Show Tags 07 Jan 2013, 11:43 1 This post received KUDOS 5C3 - select three couples 2*2*2 --> select one member from each couple ans - 5C3 * 8 = 80 Kudos [?]: -9 [1], given: 0 Manager Joined: 18 Oct 2011 Posts: 90 Kudos [?]: 91 [0], given: 0 Location: United States Concentration: Entrepreneurship, Marketing GMAT Date: 01-30-2013 GPA: 3.3 Re: GMATprep PS1 [#permalink] ### Show Tags 07 Jan 2013, 11:52 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 Kudos [?]: 91 [0], given: 0 Current Student Joined: 02 Apr 2012 Posts: 77 Kudos [?]: 59 [0], given: 155 Location: United States (VA) Concentration: Entrepreneurship, Finance GMAT 1: 680 Q49 V34 WE: Consulting (Consulting) Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 15 Jul 2013, 09:11 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 Kudos [?]: 59 [0], given: 155 Director Joined: 17 Dec 2012 Posts: 608 Kudos [?]: 515 [2], given: 16 Location: India Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 15 Jul 2013, 18:44 2 This post received KUDOS Expert's post 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? (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 http://www.sravnatestprep.com/regularcourse.php Pay After Use Standardized Approaches Kudos [?]: 515 [2], given: 16 Manager Joined: 12 Feb 2012 Posts: 130 Kudos [?]: 59 [0], given: 28 Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 21 Jul 2013, 19:38 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? Kudos [?]: 59 [0], given: 28 Math Expert Joined: 02 Sep 2009 Posts: 41871 Kudos [?]: 128513 [0], given: 12179 Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 21 Jul 2013, 22:05 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. _________________ Kudos [?]: 128513 [0], given: 12179 Intern Joined: 20 Aug 2013 Posts: 2 Kudos [?]: [0], given: 0 Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 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 Kudos [?]: [0], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7668 Kudos [?]: 17324 [0], given: 232 Location: Pune, India Re: If a committee of 3 people is to be selected from among 5 [#permalink] ### Show Tags 21 Aug 2013, 22:58 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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:

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!

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Re: If a committee of 3 people is to be selected from among 5   [#permalink] 22 Aug 2013, 05:01

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