A committee of four people

Question

A committee of four people is to be chosen from four married couples. What is the number of different committies that can be chosen if exactly one couple can serve on this committee?

A. 12×2^3
B. 12×2^2
C. 4×2^2
D. 4×2^3
E. 4×2^5

GMATGuruNY · Answer

A committee of four people is to be chosen from four married couples. What is the number of different committies that can be chosen if exactly one couple can serve on this committee?

A. 12×2^3
B. 12×2^2
C. 4×2^2
D. 4×2^3
E. 4×2^5
Let the couples be A, B, C and D
Take the task of selecting the 4 committee members and break it into stages.

Stage 1: Select one person from couple A.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 2: Select one person from couple B.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 3: Select one person from couple C.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 4: Select one person from couple D.
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 4-person committee) in (2)(2)(2)(2) ways (= 16 ways)

Answer = C

Brent,

This solution presumes that exactly ONE MEMBER of each couple must be selected.
I believe that the intended constraint is that exactly ONE COUPLE must be included on the committee.
Also, the result of this solution -- 16 -- differs from the value of answer choice C:
C: (4)(2²) = 256.

talaangoshtari · Answer

Hi GMATGuruNY,

Thank you... I saw a similar problem that you solved but order mattered. Does the order matter in this problem? What about dividing the product by 4!? I have difficulty solving these kinds of problems. What is the general approach? I have another question:

In a party of 9 people there are 2 couples. In how many ways we can select 5 people such that from each of the couples we select one person?

A. 80
B. 50
C. 60
D. 40

GMATGuruNY · Answer

talaangoshtari wrote:Hi GMATGuruNY,

Thank you... I saw a similar problem that you solved but order mattered. Does the order matter in this problem? What about dividing the product by 4!? I have difficulty solving these kinds of problems

Please revisit my post above.
I've clarified the reasoning, as follows:
Since the order of the two unmarried people does not matter -- AB is the same pair of unmarried people as BA -- we divide by the number of ways to arrange the two unmarried people (2!).

GMATGuruNY · Answer

talaangoshtari wrote::
In a party of 9 people there are 2 couples. In how many ways we can select 5 people such that from each of the couples we select one person??

A. 80
B. 50
C. 60
D. 40

Let the two couples be A and B

Take the task of selecting the 5 committee members and break it into stages.

Stage 1: Select one person from couple A.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 2: Select one person from couple B.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 3: Select three members from the remaining 5 people.
Since the order in which we select the 3 people does not matter, we can use COMBINATIONS.
We can select 3 people from 5 people in 5C2 ways (= 10 ways)

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

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus create a 5-person committee) in (2)(2)(10) ways (= 10 ways)

Answer = D

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

Brent@GMATPrepNow · Answer

talaangoshtari wrote:A committee of four people is to be chosen from four married couples. What is the number of different committies that can be chosen if exactly one couple can serve on this committee?

A. 12×2^3
B. 12×2^2
C. 4×2^2
D. 4×2^3
E. 4×2^5

Here's a slightly different approach:

Take the task of selecting the 4 committee members and break it into stages.

Stage 1: Select the couple that will serve on the committee.
There are 4 couples to choose from, so this stage can be accomplished in 4 ways.

We still need to select 2 more people, but they can't be a couple. So, we need 1 representative from 2 different couples

Stage 2: Select 2 couples from the remaining 3 couples.
Since the order in which we select the 2 couples does not matter, we can use COMBINATIONS.
We can select 2 couples from 3 couples people in 3C2 ways (= 3 ways)
So this stage can be accomplished in 3 ways.

If anyone is interested, we have a free video on calculating combinations (like 3C2) in your head: https://www.gmatprepnow.com/module/gmat ... /video/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

Last edited by Brent@GMATPrepNow on Wed Sep 16, 2015 8:36 am, edited 2 times in total.

Brent Hanneson - Creator of GMATPrepNow.com

Brent@GMATPrepNow · Answer

talaangoshtari wrote::
In a party of 9 people there are 2 couples. In how many ways we can select 5 people such that from each of the couples we select one person??

A. 80
B. 50
C. 60
D. 40

Two couples:
One person must be selected from each couple.
Number of options for the first person = 4. (Any of the 4 people who belong to a couple.)
Number of options for the second person = 2. (Either of the two people not coupled with the first person.)
To combine these options, we multiply:
4*2.
Since the order of the two people does not matter -- AB is the same pair of people as BA -- we divide by the number of ways the two people can be arranged (2!):
(4*2)/2! = 4.

Remaining people:
To complete the 5-member group, 3 people must be selected from the 5 remaining people.
From the 5 remaining people, the number of ways to choose 3 = 5C3 = (5*4*3)/(3*2*1) = 10.

To combine the options above, we multiply:
4*10 = 40.

The correct answer is D.

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GMATGuruNY · Answer

A committee of four people is to be chosen from four married couples. What is the number of different committies that can be chosen if exactly one couple can serve on this committee?

A. 12×2^3
B. 12×2^2
C. 4×2^2
D. 4×2^3
E. 4×2^5
Let the couples be A, B, C and D
Take the task of selecting the 4 committee members and break it into stages.

Stage 1: Select one person from couple A.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 2: Select one person from couple B.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 3: Select one person from couple C.
There are 2 people in the couple, so this stage can be accomplished in 2 ways.

Stage 4: Select one person from couple D.
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 4-person committee) in (2)(2)(2)(2) ways (= 16 ways)

Answer = C

Brent,

This solution presumes that exactly ONE MEMBER of each couple must be selected.
I believe that the intended constraint is that exactly ONE COUPLE must be included on the committee.
Also, the result of this solution -- 16 -- differs from the value of answer choice C:
C: (4)(2²) = 256.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

Brent@GMATPrepNow · Answer

GMATGuruNY wrote:A committee of four people is to be chosen from four married couples. What is the number of different committies that can be chosen if exactly one couple can serve on this committee? A. 12×2^3 B. 12×2^2 C. 4×2^2 D. 4×2^3 E. 4×2^5 Let the couples be A, B, C and D Take the task of selecting the 4 committee members and break it into stages. Stage 1: Select one person from couple A. There are 2 people in the couple, so this stage can be accomplished in 2 ways. Stage 2: Select one person from couple B. There are 2 people in the couple, so this stage can be accomplished in 2 ways. Stage 3: Select one person from couple C. There are 2 people in the couple, so this stage can be accomplished in 2 ways. Stage 4: Select one person from couple D. 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 4-person committee) in (2)(2)(2)(2) ways (= 16 ways) Answer = C Brent, This solution presumes that exactly ONE MEMBER of each couple must be selected. I believe that the intended constraint is that exactly ONE COUPLE must be included on the committee. Also, the result of this solution -- 16 -- differs from the value of answer choice C: C: (4)(2²) = 256. Arg! I incorrectly read the question as exactly one person from EACH COUPLE can serve! I'll edit my solution. Thanks Mitch! Cheers, Brent

Brent@GMATPrepNow · Answer

Here's a similar question to practice with: https://www.beatthegmat.com/please-solve ... 71499.html

regor60 · Answer

deleted

regor60 · Answer

My opinion is that this is a poorly worded problem, specifically, the use of the words "exactly" and "can".

To me, "can" implies " no more than", while "exactly" implies "required". To be consistent with the original answer, I think "exactly" should be paired with "will" or "shall" or "must"

The answers differ, obviously. I interpreted the question as "no more than", which allows for groups of 4 with no couples represented along with one couple, or 64 choices.

sandipgumtya · Answer

Hi mItch I didn't get the concept here.why r we dividing by 2!? Can u pl explain once more?

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