pratikbais wrote:

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

Steps to solve this problem -

People - 1M, 1F, 1S, 2D (D1,D2)

1. Choose a parent to drive the sedan.

2. Find the ways to seat the daughters.

3. Place the remaining 3 family members.

4. Finally multiply results from steps 1, 2 and 3.

1. Ways to choose a parent to drive = 2 (One person seated, total remaining = 4).

2. Arrangement in which daughters sit separate = Total Arrangements of 4 people - Arrangements with D1 and D2 glued together.

-> 4! - 4*2*1

4*2*1 - the pair of daughters can take 2 out of 3 consecutive spots at the back seat. Also they can interchange seats D1D2 and D2D1 are different arrangements. So total 4.

Remaining two people sit in 2*1 ways.

(This takes care of step 3 as well)

Finally step 4 - Total = 2* (4! - 4*2*1 ) = 2* (24-8) = 2*16 = 32.

Thanks.