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| OG - PS #195 |
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Stacey Koprince
MGMAT STAFF
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Please post the entire text including answer choices. Often, the way in which the choices are presented dictates the most effective way to solve the problem!
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Guest
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Here is the solution: (I'm just wondering if there is another way to solve the problem - an easier way) Thank you!
In order to walk from intersection X to intersection Y by one of the routes of minimum possible length, Pat must travel only upward or rightward between the intersections on the map. Let U represent upward movements and R represent rightward movements. It takes 3 upward and 2 rightward movements to complete the route. The following 10 routes are possible: UUURR UURUR UURRU URUUR URURU URRUU RRUUU RUUUR RUURU RURUU Answer is (C) where: A) 6 B) 8 C) 10 D) 14 E) 16 Thank you! |
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Guest
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1. ) Let 5! be the number of possible movements Pat can make. 5*4*3*2*1 = 120
2.) Since we need the shortest route, from looking at the grid you can see that it will take 3 upward movements and 2 movements to the right. Following the combinatory method in the book it can be written as: A B C D E U U U R R OR 3! * 2! = 12 Putting it all together you divide the first part by the second part: 5! / (3! * 2!) = 10 |
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Stacey Koprince
MGMAT STAFF
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The OG's explanation is to write it all out? Yuck! The last Guest's method does indeed show a formulaic way to approach this.
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| OG - PS #195 |
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