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If a committee of 3 people is to be selected from among 5
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Updated on: 07 Jul 2013, 13:41
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifacommitteeof3peopleistobeselectedfromamong98533.html
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Originally posted by alexandr888 on 06 Jan 2008, 02:12.
Last edited by Bunuel on 07 Jul 2013, 13:41, edited 1 time in total.
Renamed the topic, edited the question added the answer choices and OA.



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Re: A community of 3 people is to be selected ....
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Updated on: 06 Jan 2008, 08:11
10C3  5C1*8C1 = 80 Let me try to elaborate 10C3 represents we pick 3 peoples from 10 5C1*8C1 represents pick 1 couple from 5 and pick the rest one from 8 Therefore, 10C3  5C1*8C1 lead us to the result
Originally posted by Jack.Zhou on 06 Jan 2008, 07:47.
Last edited by Jack.Zhou on 06 Jan 2008, 08:11, edited 1 time in total.



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Re: A community of 3 people is to be selected ....
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06 Jan 2008, 07:54
I am not strong at that types of problems.. however I tried and got 384, am I missing some thing?



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Re: A community of 3 people is to be selected ....
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06 Jan 2008, 12:49
alexandr888 wrote: A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
(without answer choices) ABC is what we want 10!/7!*3! is number of 3 communities we can have > 120 ways. Now no married couples. Lets find the number of ways With married couples. say XXC (C stands for the other 8 possible people) we have 5 couples so 5*8=40 12040=80.



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Re: A community of 3 people is to be selected ....
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06 Jan 2008, 19:55
alexandr888 wrote: A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
(without answer choices) Hey guys, Please help to find the flaw in my reasonings ABC A: can choose 1 of 10 persons in 5 couples B: can choose 1 of 8 persons (2 persons be minus in the first couple chosen by A) C: can choose 1 of 6 persons (4 persons be minus by A and B) so total= 10*8**6=480 Your help is appreciated
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Re: A community of 3 people is to be selected ....
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06 Jan 2008, 21:27
sondenso wrote: alexandr888 wrote: A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
(without answer choices) Hey guys, Please help to find the flaw in my reasonings ABC A: can choose 1 of 10 persons in 5 couples B: can choose 1 of 8 persons (2 persons be minus in the first couple chosen by A) C: can choose 1 of 6 persons (4 persons be minus by A and B) so total= 10*8**6=480 Your help is appreciated D: N=480/3P3=480/3!=80. ABC is equal ACB, BAC, BCA, CAB, CBA
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Re: A community of 3 people is to be selected ....
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08 Jan 2008, 11:27
alexandr888 wrote: A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
(without answer choices) 10C3  5* (1*1*8c1) 120  5(8) 120  40 80
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Re: A community of 3 people is to be selected ....
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08 Jan 2008, 20:29
walker wrote: sondenso wrote: alexandr888 wrote: A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
(without answer choices) Hey guys, Please help to find the flaw in my reasonings ABC A: can choose 1 of 10 persons in 5 couples B: can choose 1 of 8 persons (2 persons be minus in the first couple chosen by A) C: can choose 1 of 6 persons (4 persons be minus by A and B) so total= 10*8**6=480 Your help is appreciated D: N=480/3P3=480/3!=80. ABC is equal ACB, BAC, BCA, CAB, CBA I think this is where Im fundamentally flawed. When do we have to do this last step of dividing by 3! ? More importantly, why are we doing it in this case ? What extra cases did we account for using the approach of 10x8x6 ? paging mr.walker, report to aisle 6



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Re: A community of 3 people is to be selected ....
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08 Jan 2008, 22:08
pmenon wrote: I think this is where Im fundamentally flawed. When do we have to do this last step of dividing by 3! ? More importantly, why are we doing it in this case ? What extra cases did we account for using the approach of 10x8x6 ? paging mr.walker, report to aisle 6 It is common for permutationcombination problems. We always have to check sensitivity to permutation. For example, we choose 3 people from 8: a) how many different placements are possible in a row: 8P3=8*7*6  XYZ is different from YXZ. b) how many different groups are possible: 8P3/3P3=8C3=8*7*6/3*2  XYZ is equivalent to YXZ.
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Re: A community of 3 people is to be selected ....
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09 Jan 2008, 05:41
walker wrote: pmenon wrote: I think this is where Im fundamentally flawed. When do we have to do this last step of dividing by 3! ? More importantly, why are we doing it in this case ? What extra cases did we account for using the approach of 10x8x6 ? paging mr.walker, report to aisle 6 It is common for permutationcombination problems. We always have to check sensitivity to permutation. For example, we choose 3 people from 8: a) how many different placements are possible in a row: 8P3=8*7*6  XYZ is different from YXZ. b) how many different groups are possible: 8P3/3P3=8C3=8*7*6/3*2  XYZ is equivalent to YXZ. when do we know that XYZ is considered different from YXZ, and when its not ? In this question, we are asked for the number of "different" groups ... is that our clue that we will need to divide by 3! in the end to account for the cases such as ZYX and XYZ, which would be considered the 'same' for this problem.



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Re: A community of 3 people is to be selected ....
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04 Sep 2009, 20:02
Walker, hats off for you and your skil with combinatorics.
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Re: A community of 3 people is to be selected ....
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06 Sep 2009, 04:16
Ans is 80.
1) The first of the 3 people group can be taken up by any of the 10 people ( 5 couples, so 5 * 2)
2) the second of the 3 people group can be taken up by any of the 8 people except the one person whose couple is already chosen as the first person.
3) The third of the 3 people group can be taken up by any of the remaining 6 people except the two people whose couples are already chosen as the first and second person.
But since the order in which people are chosen does not matter, therefore
Ans: = (10 * 8 * 6)/3! = 80



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Re: A community of 3 people is to be selected ....
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29 Oct 2009, 06:11
I am a little unclear about the order mattering. A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?Does it imply xyz is same as yzx?Or different?Please clear the doubt. Here we are dividing by 3! which means order does'nt matter??Why?
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Re: A community of 3 people is to be selected ....
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31 Oct 2009, 04:42
tejal777 wrote: I am a little unclear about the order mattering.
A community of 3 people is to be selected from 5 married couples, such that the community does not include two people who are married to each other. How many such communities are possible?
Does it imply xyz is same as yzx?Or different?Please clear the doubt. Here we are dividing by 3! which means order does'nt matter??Why? Order does not matter because this is combination ( selection ) and not permutation (arrangement ). (Mona, Tina, Mina ) and ( Tina, Mina, Mona ) is considered as 'one' community/group ( and not two different communities/groups )



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Re: A community of 3 people is to be selected ....
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31 Oct 2009, 05:08
IMO the use of the word community is what should give you the clue that order of selection does not matter. to elaborate further, since community, like the word group, implies a single entity, a community (or group) of XYZ would be the same as ZYX and so on.. think of it in this way.. you have an empty room (call it community) and you are putting people into it. after exhausting all possible cases, when you compare the rooms, you would find no difference between the room with persons XYZ and the room with persons ZYX. thats why we discount those cases and have to divide by 3!. However, suppose the room had 3 empty chairs named ABC and each person entering the room would have to sit on ABC respectively, then your single entity would become the chair and not the room. thus the situation of (XA, YB, ZC) would be different from (ZA, YB, XC). in this case, we would not have to divide by 3! and our answer would simply be 10*8*6 = 480. This is my method of understanding such problems. Let me know what you guys think.. Hope it helps!
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Re: A community of 3 people is to be selected from 5 married
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07 Jul 2013, 13:16
We have 10 people grouped in 3: 10C3 = 120 Then I substract the couples that are married, which are: AAB , AAC, AAD.. etc. For the first slot we have 5 options (the man of the couple), for the second only 1 (the woman of this man) and for the third slot, the rest of the people (8) Then: 5*1*8 = 40 120  40 = 80
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Re: A community of 3 people is to be selected from 5 married
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Re: If a committee of 3 people is to be selected from among 5
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