OG - PS #195

[Guest]

I was just wondering if there is a method to approach the following question:

(there is a 3 by 2 grid, with x at the bottom left corner, and y at the top right corner)

Pat will walk from intersection x to y along a route that is confined to the square grid of four streets and three avenues shown in the preceding map. how many routes from x to y can pat take that have the minimum possible length?

Thank you!

[StaceyKoprince]

Please post the entire text including answer choices. Often, the way in which the choices are presented dictates the most effective way to solve the problem!

[Guest]

Here is the solution: (I'm just wondering if there is another way to solve the problem - an easier way) Thank you!

In order to walk from intersection X to intersection Y by one of the routes of minimum possible length, Pat must travel only upward or rightward between the intersections on the map. Let U represent upward movements and R represent rightward movements. It takes 3 upward and 2 rightward movements to complete the route. The following 10 routes are possible:

UUURR
UURUR
UURRU
URUUR
URURU
URRUU
RRUUU
RUUUR
RUURU
RURUU

Answer is (C)

where:
A) 6
B) 8
C) 10
D) 14
E) 16

Thank you!

[Guest]

1. ) Let 5! be the number of possible movements Pat can make. 5*4*3*2*1 = 120

2.) Since we need the shortest route, from looking at the grid you can see that it will take 3 upward movements and 2 movements to the
right.

Following the combinatory method in the book it can be written as:
A B C D E
U U U R R

OR

3! * 2! = 12


Putting it all together you divide the first part by the second part:

5! / (3! * 2!) = 10

[StaceyKoprince]

The OG's explanation is to write it all out? Yuck! The last Guest's method does indeed show a formulaic way to approach this.