GMAT PREP PS question

[alex.gellatly]

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

Thanks in advanced

[killer1387]

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

Thanks in advanced

total no of ways in which committee can be formed = 8c3=56
no. of ways in which one couple is there and any one among remaining 6 is present = 4c1*6c1=24

required probability = 56-24 = 32

Hence E

[Anurag@Gurome]

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

Thanks in advanced

Number of ways to choose 3 couples from 4 couples = 4C3 = 4

Now each of these 3 couples can send two persons (husband or wife), number of ways of doing this = 2 * 2 * 2 = 2^3 = 8

Therefore, total number of ways: 4C3 * 2^3 = 32

The correct answer is E.

[GMATGuruNY]

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

Thanks in advanced

An alternate approach:

Number of choices for the 1st person = 8.
Number of choices for the 2nd person = 6. (7 people left, but we can't choose the spouse of the first committee member chosen, leaving 7-1= 6 choices.)
Number of choices for the 3rd person = 4. (6 people left, but we can't choose the spouses of the first 2 committee members chosen, leaving 6-2 = 4 choices.)

To combine the number of choices we have for each position on the committee, we multiply the numbers above:
8*6*4.

But the order in which the committee members are chosen doesn't matter: selecting ABC will yield the same committee as selecting BAC.
So that we don't overcount these duplicate combinations, we must divide by the number of ways to arrange the 3 members chosen (3!).

Thus:
Total number of possible committees = (8*6*4)/3! = 32.

The correct answer is E.