Problem Solving

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?
28
32
48
60
120

I HATE permutations and all of the sort... Just despise them... HATE probabilities too... However, I'm gonna give this one a shot just because I need to teach my brain not to go catatonic on me when it comes to these types of problems.

Here's how I see the damn thing:
First of all, you have three ways of setting up the front seats: parents, parent +girl, parent + boy.
If both parents are on the front seats, then you have two ways of arranging the back: girls either on the right or on the left. Since each parent can be the driver, you get 4 ways of arranging this freakish Adams family.
If you have a parent + girl in the front seats, then in the back you can arrange them in 3! = 6 ways. Again, since you have two parents and two girls, you get 3! * 2*2 =24 ways of arranging them.
Last but not least: if you get parent + boy, in the back they can only be arranged with both girls on the side and the other parent in the middle, which gives you 2 possible arrangements (girl A on right, girl B on left versus girl A on left and girl B on right). Since there are two parents, you get another 4 ways of setting them up.

That makes a total of 4 + 24 + 4 = 32 damn ways of having them set up...Sorry for my rant... Hope you understand smth of this rumbling...

Great explanation Dana. You Rocks.