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GMAT Set 5 Q6

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by Abhijit K » Wed Feb 18, 2015 1:13 am
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
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by GMATGuruNY » Wed Feb 18, 2015 3:36 am
Abhijit K wrote: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
Number of options for the first person = 10. (Any of the 10 people.)
Number of options for the second person = 8. (Of the 9 remaining people, anyone but the spouse of the first person selected.)
Number of options for the third person = 6. (Of the 8 remaining people, anyone but the spouses of the first two people selected.)
To combine these options, we multiply:
10*8*6.
Since the ORDER of the three people doesn't matter -- ABC is the same committee as CAB -- we divide by the number of ways the three people can be arranged (3!):
(10*8*6)/(3*2*1) = 80.

The correct answer is D.
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by Brent@GMATPrepNow » Wed Feb 18, 2015 8:55 am
Abhijit K wrote: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

Take the task of creating a committee and break it into stages.

Stage 1: Select 3 COUPLES
Since the order in which we select the couples does not matter, we can use COMBINATIONS
We can select 3 couples from 5 couples in 5C3 ways ( = 10 ways)

ASIDE: If anyone is interested, we have a free video on calculating combinations (like 5C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

At this point, we have 3 COUPLES, which we'll call A, B ans C. We're now going to select ONE person from each couple to be on the committee.

Stage 2: Select 1 person from couple A
There are 2 people in this couple, so we can complete this stage in 2 ways.

Stage 3: Select 1 person from couple B
There are 2 people in this couple, so we can complete this stage in 2 ways.

Stage 4: Select 1 person from couple C
There are 2 people in this couple, so we can complete this stage in 2 ways.

By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus create a 3-person committee) in (10)(2)(2)(2) ways ([spoiler]= 80ways[/spoiler])

Answer: D
--------------------------

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

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
- https://www.beatthegmat.com/mouse-pellets-t274303.html


MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- 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/ps-counting-t273659.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/please-solve ... 71499.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/laniera-s-co ... 15764.html

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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