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pat walking the streets

Expert replies
by djkvakin » Fri Dec 04, 2009 10:58 am
Hey, Gang - this one is from the official review 12th ed. There is an explanation in the book, but it's nothing but manual. I was wondering if anyone could come up with a faster solution. Here is the problem:
Pat is walking from Intersection X to intersection Y along the route that is confined to the square grid of four streets and three avenues shown on the map attached. How many routes from X to Y can Pat take that have a minimum possible length?

The OA is 10. In the explanation they just list the possible combinations of Upright and Right moves. There must be a mathematical solution to it.

On the attachment what i tried to accomplish was to create a 3 vertical lines that represent aveues and 4 horizontal lines that represent streets. Pat is located in the lower left corner and he has to get to the upper right. He has to walk up 3 blocks and right 2 blocks. I hope this clarifies the task. Also, for those who have OG 12th it's a math question 191.
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Source: — Problem Solving |

by Stuart@KaplanGMAT » Fri Dec 04, 2009 2:56 pm
djkvakin wrote:Hey, Gang - this one is from the official review 12th ed. There is an explanation in the book, but it's nothing but manual. I was wondering if anyone could come up with a faster solution. Here is the problem:
Pat is walking from Intersection X to intersection Y along the route that is confined to the square grid of four streets and three avenues shown on the map attached. How many routes from X to Y can Pat take that have a minimum possible length?

The OA is 10. In the explanation they just list the possible combinations of Upright and Right moves. There must be a mathematical solution to it.

On the attachment what i tried to accomplish was to create a 3 vertical lines that represent aveues and 4 horizontal lines that represent streets. Pat is located in the lower left corner and he has to get to the upper right. He has to walk up 3 blocks and right 2 blocks. I hope this clarifies the task. Also, for those who have OG 12th it's a math question 191.
Of all the math questions I get asked about in the OG, this might be the one that baffles the most students.

Here's the key: no matter which path he takes, Pat needs to go up 3 times and right twice. So, his route is going to be some permutation of Up, Up, Up, Right, Right.

Therefore, we treat this as a permutations question. We have 5 items to arrange, with a triplicate and a duplicate. So, we use the non-unique items permutations formula:

n!/r!s!t!...

in which n is the total number of items and r, s, t, etc... represent the number of duplicates.

So, the answer is 5!/3!2! = 5*4/2 = 10

That particular formula comes up most often on the GMAT in word jumble questions. The question you posted is the exact same as:

How many different ways are there to arrange the letters of the word "esses".

Some other examples of applying the formula:

How many different ways are there to arrange the letters of the word DESERT?

6!/2!

How many different ways are there to arrange the letters of the word DESSERT?

7!/2!2!

How many different ways are there to arrange the letters of the word DESSERTS?

8!/2!3!

How many different ways are there to arrange the letters of the word DEEDED?

6!/3!3!
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by djkvakin » Sat Dec 05, 2009 11:56 am
Stuart, thank you so much for your wonderful explanation. You always go above and beyond in explaining answers, and I want you to know how much we all appreciate your input.
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