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by gmatsamurai » Fri Mar 02, 2012 9:09 pm
I have tried to solve this sum, I didn't get it right. I dont really understand the explanation in the answer key. Can anybody explain?

Note: They do have a diagram, if you have your OGs it might be good idea to refer to it.

Pat will walk fro Intersection X to Intersection Y along the route that is confined to the square grid of the 4 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


Answer C
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by gmatsamurai » Fri Mar 02, 2012 9:10 pm
Thank you in advance =)
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by Anurag@Gurome » Fri Mar 02, 2012 9:18 pm
gmatsamurai wrote:I have tried to solve this sum, I didn't get it right. I dont really understand the explanation in the answer key. Can anybody explain?

Note: They do have a diagram, if you have your OGs it might be good idea to refer to it.

Pat will walk fro Intersection X to Intersection Y along the route that is confined to the square grid of the 4 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


Answer C
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.
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by GMATGuruNY » Sat Mar 03, 2012 1:35 am
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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by gmatsamurai » Sat Mar 03, 2012 2:03 am
Thanks guys =)
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