A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
a) 16
b) 24
c) 26
d) 30
e) 32.
Pls explain how to solve such problems in general
committee problem
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Number of ways to choose 3 couples from 4 couples = 4C3 = 4vishal_2804 wrote:A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
Now, each of these 3 couples can send two persons (husband or wife) for the selection of the committee.
Number of ways of doing this = 2*2*2 = 2^3 = 8
Therefore, total number of ways = 4*8 = 32
The correct answer is E.
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Here's one approach.vishal_2804 wrote:A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?
a) 16
b) 24
c) 26
d) 30
e) 32.
Pls explain how to solve such problems in general
Take the task of selecting the 3 committee members and break it into stages.
Stage 1: Select the 3 couples from which we will select 1 spouse each.
There are 4 couples, and we must select 3 of them. Since the order in which we select the 3 couples does not matter, this stage can be accomplished in 4C3 ways (4 ways)
Stage 2: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.
Stage 3: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.
Stage 4: Take one of the 3 selected couples and choose 1 person to be on the committee.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.
By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus create a 3-person committee) in (4)(2)(2)(2) ways (= 32 ways)
Answer = E
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Possibility of selecting 3 committee members where we do not want a couple to be together can be solved as:
Favorable options = Total options - options where one couple will be always there.
So total options here are 4 couples --> From 8 people select 3 -->8C3 = 56
Options where one couple will always come -->
Select 1 out of 4 couples --> 4C1 = 4 and the one position still remaining --> Select 1 from remaining 6 people = 6C1.
Therefore these options become 4 * 6 = 24
Total Options - Options with couple --> 56 - 24 = 32 ANS E
Favorable options = Total options - options where one couple will be always there.
So total options here are 4 couples --> From 8 people select 3 -->8C3 = 56
Options where one couple will always come -->
Select 1 out of 4 couples --> 4C1 = 4 and the one position still remaining --> Select 1 from remaining 6 people = 6C1.
Therefore these options become 4 * 6 = 24
Total Options - Options with couple --> 56 - 24 = 32 ANS E