In city A, the streets are aligned in a grid, where the east

Let me add a little bit of detail to the solution given above.

You need to go from a point that is to the bottom left to a point that is to the top right. So you should take steps towards right and top. Since Bill wants to take shortest possible time, he should not go left or down because that is the opposite direction. His destination lies towards right and up.
Say, he takes one step to go from one intersection to the next one. He can take various routes e.g.
RRRRRUUUU (R represents one step right and U represents one step up)
RRRRUUUUR
etc

The total number of ways is basically obtained by re-arranging 5 Rs and 4 Us. You can do it in 9!/5!*4! = 126 ways (we divide by 5! and 4! because all Rs and all Us are identical)

Now, what happens due to the park? Everything is the same except that one intersection is not available - 4th Rd, 6th Ave. You cannot include this intersection in your journey. So what do you do? You remove all paths that include this intersection.

Now our question is this: In how many ways can you go from the corner of 2nd Rd and 3rd Ave to the corner of 6th Rd and 8th Ave when you include the corner of 4th Rd, 6th Ave?
From the corner of 2nd Rd and 3rd Ave to the corner of 4th Rd and 6th Ave - You need to take 3 steps right and 2 steps up - RRRUU etc. No of ways = 5!/3!*2! = 10
From the corner of 4th Rd and 6th Ave to the corner of 6th Rd and 8th Ave - You need to take 2 steps right and 2 steps up - RRUU etc. No of ways = 4!/2!*2! = 6

Total number of ways in which the 4th Rd, 6th Ave is included = 10*6 = 60 ways

Number of paths in which the corner of 4th Rd, 6th Ave is not included = 126 - 60 = 66
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Really don't know of any formula that could help...
Pretty cruel, definitely not a real GMAT test question.

The best I could come up with is as follows:

Each optimal path (of minimal length) would consist of 5 walks to the right (R) and 4 walks up (U).
The total number of such paths is given by \(9C4=\frac{9*8*7*6}{2*3*4}=126.\)

We have to eliminate those optimal paths that go through the center of the park (the intersection between 4th Road and 6th Avenue). He can still walk around the park. To reach the center of the park we need 3R and 2U walks - a total of 5C2=5*4/2=10 possibilities. From the center of the park, to reach the final destination we need 2R and 2U walks - another 4C2=4*3/2=6 possibilities. This would give 10*6=60 paths to eliminate, leaving 126-60=66 possibilities.

Answer B (not C).

Are you sure OA is C?
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Sorry!! U are correct Ans is (B) 66. I have edited the Answer.
thank U very much.

Can u explain the logic of that 60 ways that u subtracted.

By the way OE provided as per source is below:

Draw a grid representing the problem. At every intersection of 2 lines, put the number representing how many possible ways it is to get to that intersection while walking only north and east. There's only one way of walking straight north up the most westerly avenue, so put ones all along that path. Same for the direct east path. Once the first rows, up-down and left-right, have been filled in, you can work on the internal intersections. For each intersection, add the numbers found at the intersections to the west and south of it. Since there are such and such many ways to get to the two intersections one spot away from this intersection, adding them together provides the total number of ways to get to that intersection. See below, filled in for this particular question, yielding the answer 66 possible shortest routes.

Did you understand the way I got the total number of paths 126?
The 60 is similar, but because each path has to go through the center of the park, I had to split each such path into two:
from start to the center of the park and then, from the center to destination.
So, it will be (the number of ways to reach the center) * (the number of ways to go from the center to destination).
First path is of type 3R and 2U, the second path is 2R and 2U.
Therefore, (5*4/2) * (4*3/2) = 60.

For example, for the first path, which is composed of 3 right walks and 2 up walks, the sequence is of length 5 (walks). I just have to decide out of the 5 walks, when to go up, the other three I will certainly go to the right. This is given by 5C2 = 10. This you can even check, the list is not sooo long:
UURRR
RUURR
RRUUR
RRRUU
URURR
URRUR
URRRU
RURUR
RURRU
RRURU

But it is important to understand how each path is built, and the number of choices. Then, use the appropriate formulas.
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P.S. - Didn't see you have already posted the explanation...
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Similar questions to practice:
pat-will-walk-from-intersection-a-to-intersection-b-along-a-68374.html
josh-has-to-run-an-electrical-wire-from-point-a-to-point-b-99962.html
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Center of the park is much important in this problem.
To pass through the park one must have to pass through this point and further to destination.

Ans = Total possible paths to destination(TPPD) - paths passing through the park (PPTP)

All possible paths covering min distance are permutation of pattern RRRRRUUUU R-> One step right / U-> One step up
Thus
TPPD = 9!/(5!x4!) = 126 (direct formula of permutation applied)

I have calculated PPTD in two steps first find all paths from origin to center of park:
1. Possible moves RRRUU total permutations 5!/(3!x2!) = 10
2. From park center to destination, in moves RRUU, total permutations as 4!/(2!x2!)=6.
Thus PPTD= 10 X 6 = 60

Ans = 126 -60 = 66

Refer following image to understand the logic.

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