BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Difficult Prob Question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 184
Joined: Thu Nov 25, 2010 9:57 pmFollowed by: 5 members
Thanked: 1 times
GMAT Score:700
by chaitanya.mehrotra » Mon Jul 25, 2011 10:30 am
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?
28
32
48
60
120

OA 32
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 370
Joined: Sat Jun 11, 2011 8:50 pmFollowed by: 2 members
Location: Arlington, MA.
Thanked: 27 times

by winniethepooh » Mon Jul 25, 2011 12:18 pm
The answer is 32.
You can consider simply counting with any parent driving and then adjusting the other 4 family members in 16 ways then multiply the resultant by 2 (which is for the second parent driving).

Or you can consider all ways without any restrictions then subtracting the possibilities with the sisters sitting only together.
Frong driver seat = 2 ways
Front passenger seat= 4 ways
Three back passenger seats in 3 * 2 * 1 ways.

Total = 2*4*3*2*1 = 48 ways._____(1)

Two sisters can sit together only in the back seat:
The arrsngements will now be:
Front driver seat = 2 ways.
Front passenger seat = 2 ways
Back seat can be filled by considering both daughters as 1 unit = 2c1 ways = 2 ways
As both daughters can interchange their seats = 2ways.

Total = 2*2*2*2 = 16 ways.___(2)

Number of ways with restrictions = 48-16 = 32 ways.____(1)-(2)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Thu Jul 14, 2011 7:11 amFollowed by: 1 member
Thanked: 1 times

by ECMoyano » Mon Jul 25, 2011 12:39 pm
winniethepooh wrote:The answer is 32.
You can consider simply counting with any parent driving and then adjusting the other 4 family members in 16 ways then multiply the resultant by 2 (which is for the second parent driving).

Or you can consider all ways without any restrictions then subtracting the possibilities with the sisters sitting only together.
Frong driver seat = 2 ways
Front passenger seat= 4 ways
Three back passenger seats in 3 * 2 * 1 ways.

Total = 2*4*3*2*1 = 48 ways._____(1)

Two sisters can sit together only in the back seat:
The arrsngements will now be:
Front driver seat = 2 ways.
Front passenger seat = 2 waysBack seat can be filled by considering both daughters as 1 unit = 2c1 ways = 2 ways
As both daughters can interchange their seats = 2ways.

Total = 2*2*2*2 = 16 ways.___(2)

Number of ways with restrictions = 48-16 = 32 ways.____(1)-(2)
Interesting, I did something similiar

Who can drive: 2 (mom or dad)
Who can ride shotgun: 4 (whoever isn't driving)

Than I labeled backseats A B C and gave A and C the value of the sisters.
AB, AC, BA, BC, CA, CB, and than eliminated AC + CA since they can't sit next to each other leaving me with 4 possible combinations.

2*4*4 = 32

Just wondering, how did you get Front driver and Front Passenger to be both 2?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 370
Joined: Sat Jun 11, 2011 8:50 pmFollowed by: 2 members
Location: Arlington, MA.
Thanked: 27 times

by winniethepooh » Mon Jul 25, 2011 1:32 pm
Front Driver = Mom or dad
Front Passenger: = Son or Father(This is to satisfy the condition that both sisters are sitting together, and theycan sit together only at the back seats).
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Mon Jul 25, 2011 8:37 pm
chaitanya.mehrotra 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?
28
32
48
60
120

OA 32
An alternate approach:

One daughter in the front passenger seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the front passenger seat, number of choices = 2.
Number of ways to arrange the 3 remaining people = 3! = 6.
Multiplying the results above, we get:
Number of arrangements = 2*2*6 = 24.

The two daughters separated in the back seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the leftmost back seat, number of choices = 2.
Since the 1 remaining daughter must be in the rightmost back seat, number of choices = 1.
Number of ways to arrange the 2 remaining people = 2! = 2.
Multiplying the results above, we get:
Number of arrangements = 2*2*1*2 = 8.

Thus, total arrangements = 24+8 = 32.

The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 370
Joined: Sat Jun 11, 2011 8:50 pmFollowed by: 2 members
Location: Arlington, MA.
Thanked: 27 times

by winniethepooh » Mon Jul 25, 2011 8:56 pm
Mitch can we solve it by the normal way, that is considering 4 options in the front passenger seat and then considering the restrictions at the back seat.
I understand what you've done is similar, but can we conclude it without having to add up two individual answers?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Sun Jun 13, 2010 3:07 amFollowed by: 3 members
Location: Bahrain
Thanked: 5 times
GMAT Score:720

by jainpiyushjain » Sun Mar 31, 2013 4:31 am
GMATGuruNY wrote:
An alternate approach:

One daughter in the front passenger seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the front passenger seat, number of choices = 2.
Number of ways to arrange the 3 remaining people = 3! = 6.
Multiplying the results above, we get:
Number of arrangements = 2*2*6 = 24.

The two daughters separated in the back seat:
Since one of the 2 parents must drive, number of choices for the driver's seat = 2.
Since either of the 2 daughters can be in the leftmost back seat, number of choices = 2.
Since the 1 remaining daughter must be in the rightmost back seat, number of choices = 1.
Number of ways to arrange the 2 remaining people = 2! = 2.
Multiplying the results above, we get:
Number of arrangements = 2*2*1*2 = 8.

Thus, total arrangements = 24+8 = 32.

The correct answer is B.
I get confused when to add or multiple, like for the final step in this question. Can someone please shed some light.

Thank you
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1