BTGmoderatorDC wrote:


Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length?

A. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen


OA C

Let V denote a step in the vertical direction and H denote a step in the horizontal direction. For instance, V-V-V-H-H denotes the path of walking along Avenue A until the intersection of 4th street and walking along 4th street until the point Y. Similarly, V-H-V-H-V denotes the path of walking along Avenue A, then walking along 2st street, then walking along Avenue B, then walking along 3rd street and finally walking along Avenue C to reach point Y.

We notice that a shortest path between point X and Y must include three V and two H. Further, any arrangement of three V's and two H's i.e. any arrangement of the letters V-V-V-H-H gives us a shortest path between X and Y. Using permutations with indistinguishable objects formula, we see that there are 5!/(3!*2!) = (5 x 4)/2 = 10 such arrangements. Thus, there are 10 shortest paths between points X and Y.

Answer: C

Source: Official Guide

BTGmoderatorDC wrote:


Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length?

A. Six
B. Eight
C. Ten
D. Fourteen
E. Sixteen


OA C

Source: Official Guide