GeminiHeat 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
We have - \(F\), \(M\), \(S\), \(D_1\), and \(D_2\)
Let us make the cases:
Case I:\(D_1\) and \(F\) in front with \(D_2\), \(S\) and \(M\) at the back
Possible Arrangements \(= 3! = 6\)
Case II:\(D_1\) and \(M\) in front with \(D_2\), \(S\) and \(F\) at the back
Possible Arrangements \(= 3! = 6\)
Case III:\(D_2\) and \(F\) in front with \(D_1\), \(S\) and \(M\) at the back
Possible Arrangements \(= 3! = 6\)
Case IV:\(D_2\) and \(M\) in front with \(D_2\), \(S\) and \(F\) at the back
Possible Arrangements \(= 3! = 6\)
Case V:\(S\) and \(F\) in front with \(D_1\), \(D_2\) and \(M\) at the back
Possible Arrangements \(= 2\)
Case VI:\(S\) and \(M\) in front with \(D_1\), \(D_2\) and \(F\) at the back
Possible Arrangements \(= 2\)
Case VII:\(M\) and \(F\)(driving) in front with \(D_1\), \(D_2\) and \(S\) at the back
Possible Arrangements \(= 2\)
Case VIII:\(F\) and \(M\)(driving) in front with \(D_1\), \(D_2\) and \(S\) at the back
Possible Arrangements \(= 2\)
Total possibilities \(= 6 + 6 + 6 + 6 + 2 + 2 + 2 + 2 = 32\)
Hence, option B