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Pat has to walk..

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Pat has to walk..

by Elena89 » Tue Dec 13, 2011 5:57 am
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

[spoiler]OA: C[/spoiler]
Source: OG-12

Can anyone suggest an easy way to solve this problem and others of such kind?
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by knight247 » Tue Dec 13, 2011 6:35 am
Hey Elena
All the streets and avenues are of equal length hence any route he takes will be of equal length. Its best to use the anagram method for such grid problems.

Suppose he decides the take the extreme route along the outer boundary of the grid...He has to walk
East East and then North North North. Try any random combination by urself and u'll find that every combo has two Easts and 3 Norths. So we have EENNN for short. Which can be permuted in 5!/(3!2!)=
10 Ways Hence C
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by GMATGuruNY » Tue Dec 13, 2011 6:46 am
Elena89 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

[spoiler]OA: C[/spoiler]
Source: OG-12

Can anyone suggest an easy way to solve this problem and others of such kind?
I posted a solution here:

https://www.beatthegmat.com/pat-and-his- ... cityevent=

Two similar problems:

https://www.beatthegmat.com/combinatroni ... 68412.html

https://www.beatthegmat.com/different-routes-t93698.html
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