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# In city A, the streets are aligned in a grid (see attachment

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In city A, the streets are aligned in a grid (see attachment [#permalink]

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21 Mar 2011, 20:13
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In city A, the streets are aligned in a grid (see attachment), where the east-west roads are called 1st Rd, 2nd Rd, 3rd Rd, etc, increasing in number as one moves northward. The north-south roads are called 1st Ave, 2nd Ave, 3rd Ave, etc, increasing in number as one moves eastward. There is a park that runs from 5th Ave to 7th Ave and from 3rd Rd to 5th Rd, as pictured. If Bill needs to walk from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave in the shortest possible time without walking through the park, how many different routes could he take?

1. 45
2. 54
3. 66
4. 98
5. 19

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-city-a-the-streets-are-aligned-in-a-grid-where-the-east-136949.html
[Reveal] Spoiler: OA

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Doc1.docx [19.17 KiB]

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22 Mar 2011, 19:05
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bhandariavi wrote:
In city A, the streets are aligned in a grid (see attachment), where the east-west roads are called 1st Rd, 2nd Rd, 3rd Rd, etc, increasing in number as one moves northward. The north-south roads are called 1st Ave, 2nd Ave, 3rd Ave, etc, increasing in number as one moves eastward. There is a park that runs from 5th Ave to 7th Ave and from 3rd Rd to 5th Rd, as pictured. If Bill needs to walk from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave in the shortest possible time without walking through the park, how many different routes could he take?
1. 45
2. 54
3. 66
4. 98
5. 19

There must be some short cut ways to solve this problem with combination factorial formula.
Can anyone shed light on this?

You might have seen a simpler problem without the park. You need to take shortest possible time so you cannot retrace any path. You cannot go down or left. You need to go up and right only.
To go from bottom left corner to top right, you will have to go 4 paces Up and 5 paces Right.
You can do this in many ways e.g. UUUURRRRR, UURRRUURR etc
There will be total 9!/(5!*4!) = 126 ways
Attachment:

Ques2.jpg [ 35.49 KiB | Viewed 3798 times ]

From these we need to remove those ways which use the red paths because they pass from the park. Note that you need to reach one of the thick red paths first. Only then can you reach either one of the two thin red paths. So if we just remove the cases using the thick Red paths, we will be done.

In how many ways can you go from bottom left to top right using the thick Up path?
You will need to take 3 Rights and 1 Up (in any order), then the Up thick Red path, then either (Up thin Red path, 2 Rights and 1 Up) or (Right thin Red path, 2 Ups and 1 Right)
You can do this in $$\frac{4!}{(3!)} * 1* 2 * \frac{3!}{2!} = 24$$ ways

In how many ways can you go from bottom left to top right using the thick Right path?
You will need to take 2 Rights and 2 Ups (in any order), then the Right thick Red path, then either (Up thin Red path, 2 Rights and 1 Up) or (Right thin Red path, 2 Ups and 1 Right)
You can do this in $$\frac{4!}{(2!)(2!)} * 1* 2 * \frac{3!}{2!} = 36$$ ways

Now all we need to do is: 126 - 24 - 36 = 66 ways
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22 Mar 2011, 11:29
1
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Hello)
This is not an easy question and I hope everyone not to have it on the exam)
Actually this problem is connected with wide and very interesting field of mathematics - graph theory(trees, cycles, 7 bridges of Konigsberg etc.). Probably you have heard about it. If no and you are interested - see here.
http://en.wikipedia.org/wiki/Graph_theory
However, I will try to explain it to you without any extra knowledge.

Obviously, the shortest way is to go to the right and up. I think it is easy to understand. So, the total lenght of the trip is 9 steps. Since you go from the LeftDown to RightUp corner, then you always will go through one of the "key" points, as I call them. See, i circle them in red in the picture.

Now let calculate, how many routes are there to the point 1. Obviously, there is 1 such route. I show it in yellow.

Now let calculate how many routes are there to the point 2. There are 4 such routes. You could easily understand these looking at the picture. I show these routes in yellow, green, blue and red.

How many routes are there from the origin to the 3 point? 6! See the picture.

How many routes to the 4 point? There are 4. Everything is on the picture.

And finally there is only one route to the 5 point. I hope you understand this easily.

And now let do the second part of work. Let calculate, how many options you have going from each of these 5 points to the final point - RightUp corner!
For the 1 point there is only 1 such route - going to the right.
For the 2 point there are 5 routes. See the picture, please.

From the 3 and 4 points there are 4 routes, They are quite similar, so I will show the picture only for one case.

Finally, from the point 5 there are 5 routes to the end, it is like the point 2, so I do not add the picture.

Now we are close to the answer!
What we need is to calculate how many routes, going through each of the 5 "key" points, is there. As you guess we should just multiply the number of routes from the first and second parts.
So, there are 1*1=1 routes going through point 1.
4*5=20 routes going through point 2
6*4=24 routes through point 3
4*4=16 routes through point 4
1*5=5 routes through point 5

Obviously, this is not the best or shortest way to solve this, but I hope it will help to realise better the essence of the
problem. I sorry for the "quality" of pictures, but I have not much time to draw them)
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Last edited by bagrettin on 22 Mar 2011, 21:13, edited 1 time in total.

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22 Mar 2011, 13:34
bhandariavi wrote:
In city A, the streets are aligned in a grid (see attachment), where the east-west roads are called 1st Rd, 2nd Rd, 3rd Rd, etc, increasing in number as one moves northward. The north-south roads are called 1st Ave, 2nd Ave, 3rd Ave, etc, increasing in number as one moves eastward. There is a park that runs from 5th Ave to 7th Ave and from 3rd Rd to 5th Rd, as pictured. If Bill needs to walk from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave in the shortest possible time without walking through the park, how many different routes could he take?
1. 45
2. 54
3. 66
4. 98
5. 19

There must be some short cut ways to solve this problem with combination factorial formula.
Can anyone shed light on this?

Good effort bagrettin. thanks and kudos.

After struggling a lot, this is my way of approaching this problem;

I am going to represent Avenues with suffix A and Roads with suffix R
e.g. (3A,2R) is the intersection of 3rd Avenue and 2nd Road.

Total number of shortest ways from (3A,2R) to (8A,6R) will be $$C^{9}_{5} = \frac{9!}{5!4!} = 126$$. Think it like a string "EEEEENNNN" 5 blocks east and 4 blocks north and a total of 9 blocks.

Now we need to subtract the routes that are actually not there because of the park;

We just need to consider 2 intersections for that;
(6A,3R), (5A,4R)

Ways to get to (6A,3R) from (3A,2R) = $$C^{4}_{3}=4$$
"From (6A,3R) to (6A,5R) via (6A,4R)" AND "From (6A,3R) to (7A,4R) via (6A,4R)" are two routes that need to be subtracted.
All we need to do now is to find out the number of ways to reach (8A,6R) from (6A,5R) & (7A,4R)

Ways from (6A,5R) to (8A,6R) = $$C^{3}_{2}=3$$
Ways from (7A,4R) to (8A,6R) = $$C^{3}_{1}=3$$

Number of ways to be subtracted = 4*3+4*3 = 12+12=24

Similarly, we can repeat it for intersection (5A,4R)

Ways to reach (5A,4R) from (3A,2R) = $$C^{4}_{2} = 6$$

"From (5A,4R) to (6A,5R) via (6A,4R)" AND "From (5A,4R) to (7A,4R) via (6A,4R)" are two routes that need to be subtracted.
All we need to do now is to find out the number of ways to reach (8A,6R) from (6A,5R) & (7A,4R)

Ways from (6A,5R) to (8A,6R) = $$C^{3}_{2}=3$$
Ways from (7A,4R) to (8A,6R) = $$C^{3}_{1}=3$$

Number of ways to be subtracted= 6*3+6*3 = 18+18=36

Total number of ways to be subtracted = 24+36=60

Number of possible shortest ways = 126-60 = 66.

Ans: "C"
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22 Mar 2011, 18:25
Thanks fluke & bagrettin.
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21 Aug 2011, 15:01
Thanks Karishma for this awesome explanation! I was scratching my head on a GMAT problem for the very first time in my GMAT prep. I can use this strategy in many problems Awesome!

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Re: Grid problems   [#permalink] 21 Aug 2011, 15:01
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