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If a committee of 3 people is to be selected from among 5 [#permalink]
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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
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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]
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Re: If a committee of 3 people is to be selected from among 5 [#permalink]
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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 unarrange. To arrange 3 people, you multiply by 3! To unarrange, 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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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, 11:25
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 unordered 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]
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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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Re: If a committee of 3 people is to be selected from among 5 [#permalink]
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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 \(2C1\)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 \(= 12040=80.\)
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If a committee of 3 people is to be selected from among 5 marrie [#permalink]
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07 Jan 2013, 11:43
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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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If a committee of 3 people is to be selected from among 5 [#permalink]
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23 Jan 2017, 01:14
Alexangeo wrote: 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:
\(C^5_3*(C^2_1)^3 =\frac{5*4*3}{3*2}* 2^3 = 10*8 = 80\)
(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 Hello, I have a question when you calculate the No of groups of married people you use C^2_1 but i saw a similar problem with that approach which uses C2^2 this one:If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it? when we exclude the committees with married ppl can you explain me the difference? thank you in advance! You have 5 couples. First you select 3 couples. This is done in 5C3 ways (select 3 out of 5) Next out of these 3 selected couples, from each couple, select 1 of the 2 people in 2C1 ways. Since we do it for each couple, we get 2C1 * 2C1 * 2C1 Total ways = 5C3 * 2C1 * 2C1 * 2C1 ways If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it? The same method will be used for this question as well. If there is something else done here in the solution, please send me the link of this question.
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Re: Combination/Permutation problem, couples [#permalink]
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04 Aug 2010, 04:26
Looking at the couples at first as single units was the eyeopener. Thanks very kindly.



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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, 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 unordered 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]
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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: pscombinations94068.htmlpscombinations101784.htmlcommitteeof88772.htmlif4peopleareselectedfromagroupof6marriedcouples99055.htmliftherearefourdistinctpairsofbrothersandsisters99992.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]
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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: pscombinations94068.htmlpscombinations101784.htmlcommitteeof88772.htmlif4peopleareselectedfromagroupof6marriedcouples99055.htmliftherearefourdistinctpairsofbrothersandsisters99992.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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Find the number of ways you can have a couple in the committee and then subtract from the total number of 3person 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 12040 = 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]
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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? (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]
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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?



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Re: If a committee of 3 people is to be selected from among 5 [#permalink]
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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: ifacommitteeof3peopleistobeselectedfromamong98533.html#p1051830Hope it helps.
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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



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Re: If a committee of 3 people is to be selected from among 5 [#permalink]
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21 Aug 2013, 22:58
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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