Combination math. I'm stumped. GMAT related. 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?
This means that… 
2 people (mother or father) could sit in the driver’s seat 
4 people (remaining parent or one of the children) could sit in the front passenger seat 
3 people could sit in the first back seat 
2 people could sit in the second back seat 
1 person could sit in the remaining back seat 
The total number of possible seating arrangements would be the product of these various possibilities: 2 × 4 × 3 × 2 × 1 = 48 
That's a start....
Apparently the answer is 32... but I'm not sure how to get there.
 A: This is a direct approach to the problem:
The problem has a restriction on the sister's seating and the fact only a parent can go in front. How could one sit the sisters? they can both be in the back but in the edges, or one can be co-pilot and the other one in the back.
Now look at the parents. one of them must be in the pilot, so if you say the position of the non-pilot parent and which one of the two parents are driving then you can know the position of both parents.
Case 1: one of the sisters is in front, then there are 24 combinations because you can arrange the 3 persons in the back in 6 ways and chose which sister goes front in two ways, also you can chose which parent is driving in two ways. multiplying gives you $6*2*2= 24$ ways.
case2: both sisters are in the back. Then there are 2 ways to chose which sister get which side, 2 ways to chose who drives and 2 ways to chose if the remaining parent or the son goes in front multiply them to get $2*2*2=8$ ways.
Here is an approach a la Don Larynx:
How many arrangements leave a parent in front? 2 ways to say which parent drives and then $4*3*2=24$ ways to arrange the rest gives you 48 ways.
How many arrangements have the two sisters together? one of them must be in the middle so then all that remains to be chosen is if the other sister is in the right or left (2 ways). then chose which parent drives ( 2 ways) and then chose which sister gets the middle ( 2 ways), finally chose if the parent goes front or back(2 ways). Multiply to get 16 ways that don't work. So there are $48-16=32$ that work.
A: Now subtract the ways where the daughters are next to each other. 2 choices of driver, 2 choices of daughter in the center back, 2 choices of seat for the other daughter, 2 places for brother makes $16$, leaving $32$
