GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
A family consisting of one mother, one father, two daughters and a son
[#permalink]
30 Mar 2021, 05:00
30 Mar 2021, 05:00
2
Expert Reply
8
Bookmarks
00:00
Question Stats:
64% (02:16) correct
35% (02:02) wrong based on 31 sessions
Hide Show timer Statistics
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
A. 28
B. 32
C. 48
D. 60
E. 120
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A family consisting of one mother, one father, two daughters and a son
[#permalink]
30 Mar 2021, 09:00
30 Mar 2021, 09:00
3
1
Bookmarks
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
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
Re: A family consisting of one mother, one father, two daughters and a son
[#permalink]
03 Apr 2021, 04:49
03 Apr 2021, 04:49
any shortcut for this problem
