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If a committee of 3 people is to be selected from among 5 [#permalink]
 04 Aug 2010, 04:06
Question Stats:
74% (01:21) correct
25% (00:56) wrong based on 58 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
Last edited by Bunuel on 01 Feb 2012, 14:35, edited 1 time in total.
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Re: Combination/Permutation problem, couples [#permalink]
04 Aug 2010, 04:20
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Re: Combination/Permutation problem, couples [#permalink]
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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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]
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]
29 Feb 2012, 11:24
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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Re: If a committee of 3 people is to be selected from among 5 [#permalink]
29 Feb 2012, 11:25
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]
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!
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Re: Combination/Permutation problem, couples [#permalink]
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.htmlps-combinations-101784.htmlcommittee-of-88772.htmlif-4-people-are-selected-from-a-group-of-6-married-couples-99055.htmlif-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.htmlHope 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]
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.htmlps-combinations-101784.htmlcommittee-of-88772.htmlif-4-people-are-selected-from-a-group-of-6-married-couples-99055.htmlif-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.htmlHope 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: If a committee of 3 people is to be selected from among 5 [#permalink]
27 Dec 2012, 19:27
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]
07 Jan 2013, 11:43
5C3 - select three couples
2*2*2 --> select one member from each couple
ans - 5C3 * 8 = 80
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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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