Please solve this real GMAT quant question

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The 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 serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32

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by mevicks » Wed Nov 06, 2013 12:08 am
Ankitaverma wrote:The 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 serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32
Total people = 8
Number of selections = 3

Use the slot method to find the number of ways to choose a team of 3 without including any couple:
_ _ _
_ _ 8 First person can be chosen in 8 ways
_ 6 8 Second person cannot be married to the first so 6 ways
4 6 8 Third person cannot be married to the first or second so 4 ways

Total no of ways = 8 * 6 * 4
Now to account for the repetition of 3 members divide by 3! = 32 ways

Answer E

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by [email protected] » Wed Nov 06, 2013 7:06 am
Ankitaverma wrote:The 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 serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32
mevicks' solution is great.

Here's another approach.

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)

If anyone is interested, we have a free video on calculating combinations (like 4C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by [email protected] » Wed Nov 06, 2013 7:20 am
Ankitaverma wrote:The 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 serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32
Another approach is to recognize that: # of permissible committees = total # of 3-person committees that ignore the rule - # of 3-person committees that break the rule

total # of 3-person committees that ignore the rule
If we ignore the rule about married couples, we can select any 3 people from the 8 people.
Since the order of the 3 selected people does not matter, we can use combinations.
We can select 3 people from 8 people in 8C3 ways (= 56 ways)

Aside: If anyone is interested, we have a free video on calculating combinations (like 8C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

# of 3-person committees that break the rule
We want the number of committees consisting of an entire couple and a third person.
Let's take the task of building a 3-person committee and break it into stages.

Stage 1: Select 1 of the 4 couples.
We'll place both people in this couple on the committee.
There are 4 couples, so this stage can be accomplished in 4 ways

Stage 2: Select the third person for the committee
There are now 6 people remaining, so this stage can be accomplished in 6 ways.

By the Fundamental Counting Principle (FCP) we can complete both stages (and thus create a 3-person committee) in (4)(6) ways
In other words, we can create 24 committees that break the rule.

So, # of permissible committees = 56 - 24
= 32
= E

Cheers,
Brent
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by GMATGuruNY » Wed Nov 06, 2013 8:18 am
Ankitaverma wrote:The 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 serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32
One more approach.

Number of ways to choose 3 people from 8 options = 8C3 = (8*7*6)/(3*2*1) = 56.

Determine the probability that the selected committee does not include a married couple.
Let the 3 people selected be A, B and C.
A can be any of the 8 people.
The probability that B is not the spouse of A = 6/7. (Of the 7 remaining people, 6 are not married to A.)
The probability that C is not the spouse of A or B = 4/6. (Of the 6 remaining people, 4 are not married to A or B.)
Since we want both of these events to happen, we multiply the fractions:
6/7 * 4/6 = 4/7.

Thus:
Of the 56 possible committees, the number that do not include a married couple = (4/7) * 56 = 32.

The correct answer is E.
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