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Married couple!

by gmatrant » Thu Nov 10, 2011 11:28 pm
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?

Can you please let me know what is wrong with the below approach?

Choosing the first person - 10 ways
Choosing the second person - 8 ways ( since the spouse of the first person should be ignored)
Choosing the third person - 6 ways (since the spouse of the first and second should be ignored)
10*8*6 = 480 ways
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by satishchandra » Fri Nov 11, 2011 12:13 am
gmatrant 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?
First, I will explain the solution.

We have 5 couples. Select 3 couples from them. Then select 1 member from each couple. We will have a community of 3 people in which no couple will exist.
Selecting 3 couples from 5 = 5C3
Selecting 1 person from each couple = 2c1*2c1*2c1
Total no. of ways to select = 5C3*2C1*2C1*2C1 = 80 ways

gmatrant wrote:Can you please let me know what is wrong with the below approach?

Choosing the first person - 10 ways
Choosing the second person - 8 ways ( since the spouse of the first person should be ignored)
Choosing the third person - 6 ways (since the spouse of the first and second should be ignored)
10*8*6 = 480 ways
Your approach is correct for arranging 3 member community in a row. However, this is wrong for selecting 3 member community.
The question asks us not to arrange but only to select.
This is the difference between permutations and combinations.
The no. of ways to arrange 3 members = 5p3*2c1*2c1*2c1 = 480 ways.

Take away note: Pay close attention to the terms such as select;arrange
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by neelgandham » Fri Nov 11, 2011 12:32 am
The total number of ways = 10*8*6/3! = 80.

Let us take an example. Let A,a ; B,b ; C,c ; D,d ; E,e be the couples.The communities formed by just three people A,B,C are

A-B-C
C-B-A
C-A-B
A-C-B
B-A-C
B-C-A

However, Order is not important in this case. So, A-B-C = C-B-A and so on...For each set of three individuals the the same community arrangement repeats 6 times. So, in your case, you have left the problem half solved. Hope it helps !
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by shankar.ashwin » Fri Nov 11, 2011 12:42 am
In some cases if you're not sure, just consider the number of ways you select 3 from 10, that is 10C3 = 120.

So you know your answer is wrong here, it should obviously be lesser than 120.
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by user123321 » Fri Nov 11, 2011 5:05 am
you can also subtract the combination formed with couples from total number of possible combinations.

selecting 3 people from 10 involves 10C3 combinations.
if a couples are present among these three people. then two persons should be a couple and other a single. selecting a couple = 5c1. selecting third person from remaining people = 8c1
and possibilities = 8*5 = 40

so communities possible = 120-40 = 80

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