Married couple: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. 32
B. 48
C. 60
D. 72
E. 80
Please revisit my post above.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
Two couples: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
Here's a slightly different approach: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
Let the two couples be A and Btalaangoshtari 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
Brent,Let the couples be A, B, C and DA 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
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
Arg! I incorrectly read the question as exactly one person from EACH COUPLE can serve!GMATGuruNY wrote:Brent,Let the couples be A, B, C and DA 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
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
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.
