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 →

PS | OG 12 | Combinatorics

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 206
Joined: Sun Jun 24, 2012 5:44 pmFollowed by: 3 members
Thanked: 5 times
by [email protected] » Sat Oct 06, 2012 1:49 pm
Pat will walk from Intersection X to Intersection Y
along a route that is confined to the square grid of
four streets and three avenues shown in the map
above. How many routes from X to Y can Pat take
that have the minimum possible length?

(A) 6
(B) 8
(C) 10
(D) 14
(E) 16

Refer page 179 for diagram. Can the answer be arrived with a formula or a combinations strategy rather than counting? Please help.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 279
Joined: Mon Jun 25, 2012 10:56 pmFollowed by: 10 members
Thanked: 60 times

by anuprajan5 » Sat Oct 06, 2012 2:04 pm
Hey Nishat,

It would be great if I could get the image. But off the top of my head, this can be done with a combinations strategy.

Its a strategy which uses how many routes we can create from recurring items. For example, in this problem if we had a grid which showed 4 downs and 4 rights from X to Y, then we would find the number of arrangements by

(Total arrangements!)/ ((down!)*(right!))



Regards
Anup
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Sun Oct 07, 2012 8:11 pm
[email protected] wrote:Pat will walk from Intersection X to Intersection Y
along a route that is confined to the square grid of
four streets and three avenues shown in the map
above. How many routes from X to Y can Pat take
that have the minimum possible length?

(A) 6
(B) 8
(C) 10
(D) 14
(E) 16

Refer page 179 for diagram. Can the answer be arrived with a formula or a combinations strategy rather than counting? Please help.
Image

For the length to be minimum, Pat should eight go upwards or right. So, for this he goes 3 steps up and then 2 steps right or 2 steps right and then 3 steps up, which makes 5 steps in all.
So, number of routes from X to Y that Pat can take having the minimum possible length = 5C2 = 5!/(3!2!) = 10

The correct answer is C.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1