Problem Solving

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

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OA: C[/img].

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?
45
54
66
98
19

Good routes = total routes - bad routes through the park.

Let E = each block traveled eastward and N = each block traveled northward.

Total routes:


To travel from A to F, 5 blocks must be traveled eastward and 4 blocks must be traveled northward.
Let the 5 blocks traveled eastward = EEEEE.
Let the 4 blocks traveled northward = NNNN.
Any arrangement of EEEEENNNN will yield a possible route, since any arrangement of these letters will represent exactly 5 blocks traveled eastward and 4 blocks traveled northward.
Number of ways to arrange EEEEENNNN = 9!/5!4! = 126.

Bad routes:
A bad route goes through the park.
There are two entryways into the park: point B and point E.

Through point B:


Bad route A-B-C-F:
To travel from A to B, 2 blocks must be traveled eastward (EE) and 2 blocks must be traveled northward (NN).
Number of ways to arrange EENN = 4!/2!2! = 6.
Number of ways to travel through the park from B to C = 1.
To travel from C to F, 2 blocks must be traveled eastward (EE) and 1 block must be traveled northward (N).
Number of ways to arrange EEN = 3!/2! = 3.
To combine the options above, we multiply:
6*1*3 = 18.

Bad route A-B-D-F:
As shown above, the number of ways to travel from A to B = 6.
Number of ways to travel through the park from B to D = 1.
To travel from D to F, 1 block must be traveled eastward (E) and 2 blocks must be traveled northward (NN).
Number of ways to arrange ENN = 3!/2! = 3.
To combine the options above, we multiply:
6*1*3 = 18.

Through point E:


Bad route A-E-C-F:
To travel from A to E, 3 blocks must be traveled eastward (EEE) and 1 block must be traveled northward (N).
Number of ways to arrange EEEN = 4!/3! = 4.
Number of ways to travel through the park from E to C = 1.
As shown above, the number of ways to travel from C to F = 3.
To combine the options above, we multiply:
4*1*3 = 12.

Bad route A-E-D-F:
As shown above, the number of ways to travel from A to E = 4.
Number of ways to travel through the park from E to D = 1.
As shown above, the number of ways to travel from D to F = 3.
To combine the options above, we multiply:
4*1*3 = 12.

Good routes = 126-18-18-12-12 = 66.

The correct answer is C.

Thanks Mitch. Simple and clean solution as always.

i did not get the explanation for calculation of bad paths....need some help here.

1947 wrote:i did not get the explanation for calculation of bad paths....need some help here.

I've added a bit more detail to my post above.

Thanks a lot Mitch for explaining so patiently....makes sense now.

GMATGuruNY wrote:

1947 wrote:i did not get the explanation for calculation of bad paths....need some help here.

I've added a bit more detail to my post above.

Great explanation Mitch.
The trick is to find number of bad routes. I found given method for finding bad routes bit difficult.
I did it as below:

To follow a bad route Bill must pass through centre of the park.

As explained by Mitch above, there are only to entries (B and E) to the park.
Total number of path to arrive at centre of park via B = (4!)/(2!)(2!) = 6
Total number of path to arrive at centre of park via E = (4!)/(3!) = 4

Total number of ways ways Cto arrive at centre of park via B or E = 6+4 = 10
Once Bill s at centre of park, he can go to F in (4!)/(2!)(2!) = 6 ways

So total number of bad routes from A to F = 10*6 = 60
Good routes = Total routes - bad routes
= 126 - 60 = 66 i.e C
Mitch: Do you see any issue with the approach that I followed?

Just curious ...Is this a real GMAT question ?