gmatnmein2010 wrote:1. 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


i took a lot of time and cud not solve it
pls suggest a short method
thanks

Case 1:-Only 1 parent driving+ 1 daughter in front seat . 2C1 x 2C1(selection of front seat daughter) x 3!(back seat arrangement)=24
Case 2 :-Both parents in the front seat. 2C1(for driver) x 2!(daughters arrangementson always sits in middle)=4
Ans. is 28.A

Total ways to seat them with one parent driving: 2 options for driver * 4 options for other seats = 2*4*3*2*1 = 48.
Assuming the 2 sisters do sit together, front seat we have 2 options for driver and 2 options for passenger (the other parent + son) = 4 ways.
Back seats: sisters sit together so either they sit in seats 1 and 2 or seats 2 and 3 = 2 ways. Also sisters themselves can sit in 2 ways so total = 2*2 = 4. So total ways sisters can sit together = 4*4 =16.
So ways they wont sit together = 48-16 = 32 ways.

gmatnmein2010 wrote:1. 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


i took a lot of time and cud not solve it
pls suggest a short method
thanks

there can be two cases
Case 1: Only 1 parent driving + 1 daughter in front seat .
No. of combination= 2C1 x 2C1 x 3!=24
Case 2 : Both parents in the front seat. and in back seat girls will at extreme and son in between = 2C1 x 2!=4
Ans. is 28A