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dunkin77: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 01, 2007 5:41 am

OG question

by dunkin77 » Sat Apr 28, 2007 9:00 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

Hi,

I solved this question by counting each way,,,,but I think there should be easier or effective way to do it. Can anyone please help?

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Source: Beat The GMAT — Problem Solving | Sat Apr 28, 2007 9:00 am

Cybermusings: Legendary Member; Posts: 559; Joined: Tue Mar 27, 2007 1:29 am; Thanked: 5 times; Followed by:2 members

by Cybermusings » Sun Apr 29, 2007 3:06 am

Whats the OA?

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Princeton Review: GMAT Instructor; Posts: 22; Joined: Fri Mar 16, 2007 10:28 pm

by Princeton Review » Sun Apr 29, 2007 3:41 am

Counting each way is perfectly acceptable. I don't know how comfortable you are with the math, but for if you're a mid-scoring student I would probably tell you to count 'em up.

To get there in the shortest route, Pat has to go right two and up three. Let's call those RR and UUU.

Since there are 5 moves pat has to make, you would start out with 5!. But some of those moves are the same as each other (e.g. the U's). Since it doesn't matter which of the U's he does at a particular moment, you have to divide out to get rid of the double-counting. So the equation would look like

5!/[(2!)*(3!)]

The 2! and 3! are dividing out the repetition of the RR and UUU respectively.

Hope that makes sense. This is one of the complex math concept o th GMAT, so it can be a little tricky to get from the boards.

M

Matt McIver

Princeton Review Instructor