A committee of 3 people is to be formed from 4 married couples. What is the number of different committees that can be chosen if 2 people who are married to each other cannot both serve on the committee ?
(AA1) (BB1) (CC1) (DD1)
The first person can be selected 8 different ways.
The second person can be selected 6 different ways. (if you selected A, you can't select A1).
The third person can be selected 4 different ways (if you selected B in second attempt , you can't select B1)
ABC1 and C1AB are same since order doesn't matter.
Therefore number of committees = (8*6*4)/3! =32
(E) is my answer.
"Education is what remains when one has forgotten everything he learned in school."