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

Another way to think about this problem:

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) to send only one "representatives" to the committee: 5C3=10.

But these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Total # of ways: 5C3*2^3=80.

Answer: D.
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total no for selecting 3 out of 10=10c3=120

no. of ways in which no two married people included= tot- 2 married couple included
2 married couple can be included in 5c1( no. of ways selecting a couple) * 8c1( no. of ways selecting the third person)=5 * 8=40

reqd comb=120-40=80
hence D

I too got 80 with the conventional way of 10C3 - 5C1 * 8C1 = 120 - 40 = 80.
But learnt and loved Bunuel's way. Thanks!

Looking at the couples at first as single units was the eye-opener. Thanks very kindly.

5C3 --> ways 3 couples can be selected out of 5 couples = 10
(2C1)^3 --> ways to select 1 adult from each couple, for each of the three couples. = 8
Total Possibilities = 5C3 * (2C1)^3 = 80

You have to select a committee of 3 people. It means A, B, C form the same committee as B, C, A i.e. you don't have to arrange people. But when you say 'first person' can be selected in 10 ways, you are arranging the 3 people in first, second and third spots. So according to you, A, B, C and B, C, A are different committees.
So all you need to do in un-arrange. To arrange 3 people, you multiply by 3!
To un-arrange, you will divide by 3!
480/3! = 80

There's your answer.

Check out this post for a detailed explanation:
http://www.veritasprep.com/blog/2011/11 ... nstraints/
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How it could be two person, out of three couples they have to sent 3 rep rite.. I am totally confused kindly enlighten me

I always use the ANAGRAM technique.

How many ways can we select three persons from 5 couples? Since we can only get one representative from each couple selected, we can imagine the 5 couples as 5 persons.

5!/3!2! = 10

Now, we know that there are 10 ways to select three persons of representing couples. But whenever we make selections, there are two ways to select from the couple, either wife or husband. Thus, 2 x 2 x 2 = 8

8*10 = 80

Answer: C

Find the number of ways you can have a couple in the committee and then subtract from the total number of 3-person committees.
10C3 = 120 possible combinations
Since there are 5 couples and the final spot could be filled with any of the 8 remaining people you have 8x5 = 40 ways to achieve this.
Therefore you have 120-40 = 80 committees w/out a married couple. Answer: D

i.e., IF AB, CD, EF, GH, IJ are the couples you have to select only one from a group

1. So selecting one from a group of 2 can be done in \(2C1\)ways. i.e., in 2 ways You have to select 3 people that way. So the total number of possibilities is \(2*2*2 = 8\)

2. Each group of 2 itself has to be selected from 5 such groups. You are selecting 3 groups of 2 from 5 such groups. therefore the total number of possibilities for this is \(5C3= 10\).

3. \((1) * (2) = 8* 10 = 80.\)

1. Total number of possibilities of selecting 3 people out of 10 people\(= 10C3= 120\)
2. For total number of groups with a married couple the situation is you have (i)2 people who are married to each other i.e., a group of 2 married to each other and (ii) one other person. The number of possibilities of (i) is the number of ways one group of married people can be selected from 5 groups of married people which is \(5C1 = 5\)ways, the number of possibilities of (ii) can be arrived by finding out in how many ways the remaining person can be selected. It can be done in \(8C1=8\) ways because if you remove the selected married couple 8 persons will remain.
3. Total number of ways of having a married couple\(in the group of 3 = 5*8=40\)4. So number of groups in which 2 people are not married couple \(= 120-40=80.\)
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