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PS | OG 12 | Combinatorics

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[email protected]: 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.

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Source: Beat The GMAT — Data Sufficiency | Sat Oct 06, 2012 1:49 pm

anuprajan5: 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

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Anurag@Gurome: 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.

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)

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