Combination problem

The posted problem is the same as the following:

The 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 serve on the committee?

a. 16
b. 24
c. 26
d. 30
e. 32

Number of options for the 1st person = 8.
Number of options for the 2nd person = 6. (Of the 7 people left, we can't use the mate of the 1st person chosen, leaving 7-1= 6 choices.)
Number of options for the 3rd person = 4. (Of the 6 people left, we can't use the mates of the 2 people already chosen, leaving 6-2 = 4 choices.)

Since the ORDER of the selections doesn't matter -- ABC is the same COMMITTEE as BCA -- we divide by the number of ways to arrange the 3 people chosen:
(8*6*4)/3! = 32.

The correct answer is E.

One more approach:

Number of ways to choose 3 people from 8 options = 8C3 = (8*7*6)/(3*2*1) = 56.

Determine the probability that the selected committee does not include a married couple.
Let the 3 people selected be A, B and C.
A can be any of the 8 people.
The probability that B is not the spouse of A = 6/7. (Of the 7 remaining people, 6 are not married to A.)
The probability that C is not the spouse of A or B = 4/6. (Of the 6 remaining people, 4 are not married to A or B.)
Since we want both of these events to happen, we multiply the fractions:
6/7 * 4/6 = 4/7.

Thus:
Of the 56 possible committees, the number that do not include a married couple = (4/7) * 56 = 32.

The correct answer is E.

A third 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.