one of the two parents can sit in front in 2 ways
two girls do not sit together. so lets find the case when they do sit together and we subtract it from total.
so arrangements in which the two girls sit together = always in the back seat of the car with one more person (either of parents or the son). this can happen in 2*2 ways = 4 ways
now the 2nd front seat of the car can be filled in 2 ways (either one of the parent or the son)
hence total ways in which atleast one parent is in front and 2 girls always sit together = 2*2*2*2 = 16 ways
total ways only when 1 parent sits in front (no other condition) = 2*4! = 48 ways

required number of ways = 48-16 = 32 ways
hence B

manish.jaipatna wrote:Hi All,

Can you please help me in solving below problem.

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?

A. 28
B. 32
C. 48
D. 60
E. 120

driver's seat can be filled in 2 ways (as one of the parents must drive);

seat adjacent to driver's seat can be filled by one of the parent, son, one of the two daughters.
lets call one of the parent as p1, son as s, daughters as d1 and d2;

when the d1 sits at the adjacent position to driver's seat then the remaining 3 persons can arrange themselves in the back seat in 3! ways i.e. 6; similarly for d2 we have 6 ways;

when the p1 sits the adjacent position to driver's seat then , son must sit in the center to make sure that daughters don't sit next to each other d1 s d2..!! d1 and d2 can interchange their positions in 2 ways, so therefore total no. of arrangements are 2

similarly when son sits at the seat adjacent to driver's seat total no. of ways of arrangements would be 2.!!

hence the required arrangement is 2(6+6+2+2)=2*16 = 32; hence B