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

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 !

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.

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

user123321