Hi All,
This question is fairly high-concept, but the math behind it isn't too complex. You have to be organized and you have to remember that we're forming GROUPS of people, so 'duplicate entries' are not allowed.
From the prompt, we have 4 married couples, but we have to form a group of 3 without putting a married couple in the group. Let's work through the logic in pieces...
The 1st person in the group can be ANY of the 8 people. Once we pick one of those 8, then we CANNOT pick the married partner to that person...
For the 2nd person in the group, we now have 6 people to choose from. Once we pick one of those 6, then we CANNOT pick the married partner to that person...
For the 3rd person in the group, we now have 4 people to choose from.
We now multiply those three numbers together: (8)(6)(4) = 192
We're NOT done though. Remember that we're forming GROUPS of people, and the 192 options we've figured out so far contain LOTS of duplicates... If you have 3 people: A, B and C, then that is just ONE group. However, in the above approach, you can end up with that group in 6 different ways:
ABC
ACB
BAC
BCA
CAB
CBA
Thus, every actual option has been counted 6 times, so we have to divide that total by 6...
192/6 = 32
Final Answer:
GMAT assassins aren't born, they're made,
Rich