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If a committee of 3 people is to be selected from among 5 [#permalink]

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06 Jan 2008, 03:12

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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?

Re: A community of 3 people is to be selected .... [#permalink]

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06 Jan 2008, 08:47

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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

Last edited by Jack.Zhou on 06 Jan 2008, 09:11, edited 1 time in total.

Re: A community of 3 people is to be selected .... [#permalink]

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06 Jan 2008, 13:49

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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

Re: A community of 3 people is to be selected .... [#permalink]

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06 Jan 2008, 20:55

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

Re: A community of 3 people is to be selected .... [#permalink]

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06 Jan 2008, 22:27

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Expert's post

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

Re: A community of 3 people is to be selected .... [#permalink]

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08 Jan 2008, 12:27

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?

Re: A community of 3 people is to be selected .... [#permalink]

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08 Jan 2008, 21:29

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 ?

Re: A community of 3 people is to be selected .... [#permalink]

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08 Jan 2008, 23:08

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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 permutation-combination 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. _________________

Re: A community of 3 people is to be selected .... [#permalink]

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09 Jan 2008, 06:41

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 permutation-combination 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.

Re: A community of 3 people is to be selected .... [#permalink]

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06 Sep 2009, 05:16

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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

Re: A community of 3 people is to be selected .... [#permalink]

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29 Oct 2009, 07:11

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? _________________

Re: A community of 3 people is to be selected .... [#permalink]

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31 Oct 2009, 05:42

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 )

Re: A community of 3 people is to be selected .... [#permalink]

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31 Oct 2009, 06:08

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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 (X-A, Y-B, Z-C) would be different from (Z-A, Y-B, X-C). 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! _________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

Re: A community of 3 people is to be selected from 5 married [#permalink]

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07 Jul 2013, 14:16

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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 _________________

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

Re: A community of 3 people is to be selected from 5 married [#permalink]

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07 Jul 2013, 14:40

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Maxirosario2012 wrote:

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

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.

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