205. A committee of three people is to be chosen from four m

[varun289]

205. 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

[Jim@StratusPrep]

8 x 6 x 4 /(3 x 2 x 1) = 32

There are 8 people available for the first of three spots, then only six because both members of the couple that had a member chosen are eliminated. The same thing happens to move from 6 to 4 for the final spot.

You divide by 3! or 3 x 2 x 1 because this is a group and order does not matter.

[GMATGuruNY]

205. 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

An alternate approach:

Select 3 couples:
From the 4 couples, the number of ways to choose 3 = 4C3 = (4*3*2)/(3*2*1) = 4.

From each of the 3 couples, select one member:
Number of options from the 1st couple = 2.
Number of options from the 2nd couple = 2.
Number of options from the 3rd couple = 2.
To combine these options, we multiply:
2*2*2 = 8.

To combine the options above, we multiply:
4*8 = 32.

The correct answer is E.

[Anindya Madhudor]

Here is an alternative way of doing this.

Number of committees with non-married people = Total number of committees - Number of committees with married couple.

Total committee= 8C3=56

When you have the first married couple in the committee, you can choose any of the other 6 members to form the committee. So, a committee with the first couple can be formed in 6 ways. Similarly, committees with the second or third or fourth couple can be formed in 6 ways.

Total number of committees with married couple = 6*4 = 24

Number of committees with non-married people = 56 -24 =32

[puneetkhurana2000]

Total Ways = 8C3 = 56

Number of ways to select a married couple are = 4C1 (1 married couple from 4 couples) * 6C1 (1 from rest 6) = 24

Remaining are 56 - 24 = 32.

[Brent@GMATPrepNow]

205. 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

Here's one 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 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

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this free video: https://www.gmatprepnow.com/module/gmat- ... /video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Then you can try solving the following questions:

EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
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- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html


DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

[Brent@GMATPrepNow]

205. 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

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