Sorry if this was already posted but I could not find it....
A committee of 3 people is to be choosen from 4 married couples. What is the # of different committees that can be formed if 2 people married to eachother can't both serve on the same commeettee?
My approach was :
1st place = 8 people
2nd place = 6 people
3rd place = 4 people
answer: 8x6x4. why is this wrong?
Only one of the four couples will not have a member of the committee, so there are 4 ways to choose which couple will be excluded. Then each of the 3 couples represented will choose which member to put on the committee: (2*2*2) ways. So in total, we have 4*8=32