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A committee of 3 people is to be chosen from four married [#permalink]
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11 May 2010, 12:35
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A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee? A. 16 B. 24 C. 26 D. 30 E. 32 Approach I thought is as follows...if some shorter method is possible please explain..
total selections = 8C3 = 56
let's say that couple is always present in this committee of three.
This means that there are 4 ways to select 2 people of the committee. ( 4 couples and any one couple can be selected in 4 ways) The third person can be selected out of remaining 6 people in 6 ways.
Therefore when couple exists there are: 4X6 = 24 ways
Thus no couple = 8C3  (4X6) = 32
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Last edited by Bunuel on 27 Oct 2013, 05:44, edited 2 times in total.
Edited the question and added the OA



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Re: PSCombinations [#permalink]
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11 May 2010, 15:24
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Re: PSCombinations [#permalink]
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Hi everybody, I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.
i can choose first member of the comitee in 8 ways.. removing the spouse of the selected person second member can be chosen in 6 ways.... third member in 4 ways..... so 8*6*4 which is not answer can someone explain why?



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Re: PSCombinations [#permalink]
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amitjash wrote: Hi everybody, I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.
i can choose first member of the comitee in 8 ways.. removing the spouse of the selected person second member can be chosen in 6 ways.... third member in 4 ways..... so 8*6*4 which is not answer can someone explain why? The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people  3! > 192/3!=32. Consider this: there are two couples and we want to choose 2 people not married to each other. Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible: \(A_1,B_1\); \(A_1,B_2\); \(A_2,B_1\); \(A_2,B_2\). Only 4 such committees are possible. If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! > 8/2!=4. You can see the similar problem at: committeeof88772.html#p669797Hope it helps.
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Re: PSCombinations [#permalink]
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22 Jul 2010, 08:37
Bunuel wrote: The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people  3! > 192/3!=32.
Thanks the two of you! This is also the way I like to solve such questions and I believe it is way faster than any 10C3... and so on!



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Re: permutation problem [#permalink]
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28 Oct 2010, 22:45
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So there are 3 people we need to choose from 8. Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways Total ways = 4+4+12+12 = 32 Posted from my mobile device
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Re: permutation problem [#permalink]
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29 Oct 2010, 02:28
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Since there are 4 couples so we have 8 people involved. The First person can be selected from the 8 people in 8 ways The second person should not be a spouse of the first and hence we have 6 ways to choose him/her The Third person should not be a spouse of either of the 2, so we can choose him in 6 ways. So the total no. of ways we can choose the people will be 8*6*4 ways. However since order is not important (i.e A,B,C is the same as B,A,C) so we divide the total ways by 3! Hence the total number of groups is 32
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Re: Combination [#permalink]
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Re: PSCombinations [#permalink]
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25 Apr 2011, 04:58
4C3 = 4 (# of ways couples can be chosen) => total # of ways = 4 * 2 * 2 * 2 (2 options from each couple) = 32 Answer  E
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Re: permutation problem [#permalink]
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09 Apr 2012, 14:27
devashish wrote: Since there are 4 couples so we have 8 people involved.
The First person can be selected from the 8 people in 8 ways The second person should not be a spouse of the first and hence we have 6 ways to choose him/her The Third person should not be a spouse of either of the 2, so we can choose him in 6 ways.
So the total no. of ways we can choose the people will be 8*6*4 ways. However since order is not important (i.e A,B,C is the same as B,A,C) so we divide the total ways by 3!
Hence the total number of groups is 32 Hey, can you help me how you get the 3!. What does it stand for or what does this number say? Thanks



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Re: permutation problem [#permalink]
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10 Apr 2012, 04:31
andih wrote: devashish wrote: Since there are 4 couples so we have 8 people involved.
The First person can be selected from the 8 people in 8 ways The second person should not be a spouse of the first and hence we have 6 ways to choose him/her The Third person should not be a spouse of either of the 2, so we can choose him in 6 ways.
So the total no. of ways we can choose the people will be 8*6*4 ways. However since order is not important (i.e A,B,C is the same as B,A,C) so we divide the total ways by 3!
Hence the total number of groups is 32 Hey, can you help me how you get the 3!. What does it stand for or what does this number say? Thanks It seems that you need to brush up your fundamentals: Factorial: http://mathworld.wolfram.com/Factorial.htmlCombinatorics: mathcombinatorics87345.htmlHope it helps.
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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06 Aug 2012, 11:21
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Person (p1 p2 p3 p4 p5 p6 p7 p8) No of ways to choose 1st Person: Any 8 No of ways to choose 2nd Person: 6 (Pair of 1st person can not be considered so we need to exclude 1 pair) No of ways to choose 3rd Person: 4 (Pair of 1st & 2nd Person can not be considered so we need to exclude 2 pair) No of ways : 8X6X4 (Now we have done a permutation) But here order of the team member is not important and 3 person can arrange themselves in 3! ways. So need to divide the permutation by 3!. Ans: 8*6*4/3! = 32
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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06 Aug 2012, 14:14
The method I used to avoid combinations
Line up the people: 12 34 56 78
Then we have something like this: 135 136 137 138 145 146 147 148
So for the "1's" we have 8 numbers. For "2" we will also have 8, and another 8 more for both "3" and "4".
8 * 4 = 32



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Re: permutation problem [#permalink]
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05 Nov 2012, 23:28
shrouded1 wrote: So there are 3 people we need to choose from 8.
Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways
Total ways = 4+4+12+12 = 32
Posted from my mobile device not sure whether this is an easy way.. but i understood this quite well except the part marked in red. is that like if 2 men has been already selected from 2 couples we are looking for women in two ways from other two couples.



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Re: permutation problem [#permalink]
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06 Nov 2012, 03:47
breakit wrote: shrouded1 wrote: So there are 3 people we need to choose from 8.
Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways
Total ways = 4+4+12+12 = 32
Posted from my mobile device not sure whether this is an easy way.. but i understood this quite well except the part marked in red. is that like if 2 men has been already selected from 2 couples we are looking for women in two ways from other two couples. Exactly. If we choose 2 men out of 4, then the third person must be a woman from the remaining two couples: 4C2*2=6*2=12. Hope it's clear.
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Re: permutation problem [#permalink]
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06 Nov 2012, 07:23
Bunuel wrote: breakit wrote: shrouded1 wrote: So there are 3 people we need to choose from 8.
Case 1 : all 3 are men ... C(4,3)=4 ways Case 2 : all 3 are women ... C(4,3)=4 ways Case 3 : 2 men 1 woman ... Choose men in C(4,2) ways, then we can only choose the woman in 2 ways, since their wives can't be chosen .., hence 12x2=24 ways Case 4 : 2 women 1 man ... Exactly same logic as case 3, 12 ways
Total ways = 4+4+12+12 = 32
Posted from my mobile device not sure whether this is an easy way.. but i understood this quite well except the part marked in red. is that like if 2 men has been already selected from 2 couples we are looking for women in two ways from other two couples. Exactly. If we choose 2 men out of 4, then the third person must be a woman from the remaining two couples: 4C2*2=6*2=12. Hope it's clear. 4C2*2=6*2=12. ??? (4*3*2*1)/(2*1) = 12 but you have mentioned as 6.. is something I am missing here.



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Re: permutation problem [#permalink]
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Re: permutation problem [#permalink]
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06 Nov 2012, 12:01
Bunuel wrote: breakit wrote: 4C2*2=6*2=12. ??? (4*3*2*1)/(2*1) = 12 but you have mentioned as 6.. is something I am missing here. \(C^2_4=\frac{4!}{2!2!}=6\). \(C^2_4=\frac{4!}{2!2!}=6\).[/quote] sorry , my mind is totally screwed with above stuff.. help me here



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Re: permutation problem [#permalink]
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Re: A committee of 3 people is to be chosen from four married [#permalink]
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06 Nov 2012, 12:53
is this a correct way to get the answer? or was it just coincidence:
8C3  4(4C2) = 56  4(6) = 32




Re: A committee of 3 people is to be chosen from four married
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