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Junior | Next Rank: 30 Posts
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Joined: Mon Aug 22, 2011 1:56 am
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by Syleena » Wed Sep 14, 2011 1:30 am
Can anyone please share a quicker way to solve this:

Pat will walk from X to Y along a route that is confined to the square grid of four streets and three avenues shown. How many routes from X to Y can Pat take that have minimum possible lenght?
A.6
B.8
C.10
D.14
E.16

The OG 12's solution is to list allthe routes with 3 upward and two rightward movements and count them. The correct answer is C. In my opinion this solution takes more than 2 minutes. Does anyone know a faster way to solve this ???
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by Geva@EconomistGMAT » Wed Sep 14, 2011 1:43 am
Syleena wrote:Can anyone please share a quicker way to solve this:

Pat will walk from X to Y along a route that is confined to the square grid of four streets and three avenues shown. How many routes from X to Y can Pat take that have minimum possible lenght?
A.6
B.8
C.10
D.14
E.16

The OG 12's solution is to list allthe routes with 3 upward and two rightward movements and count them. The correct answer is C. In my opinion this solution takes more than 2 minutes. Does anyone know a faster way to solve this ???
Conceptually more difficult, but less time consuming: the question is asking "how many ways are there to arrange 3 Us and 2 Rs", where a U means "go up" and an R means "go Right".

the number of ways of ordering five different letters is 5!. But here the 3 Us are the same (internal order does not matter), and the two Rs are the same (internal order does not matter). In such cases, divide by the size of the internal group factorial. the result is 5!/3!2! = 5*4/2 = 10

Note that this calculation is the same a 5C3, which presents another way to look at the problem: find the number of ways to place 3 Us in five places (choose 3 out of 5, order does not matter), and the Rs will fall into place in the remaining two.
Geva
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Junior | Next Rank: 30 Posts
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by Syleena » Thu Sep 15, 2011 11:42 pm
Thank you! :)
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