Problem Solving

OG 11, problem 195 on page 178.

Q: 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 (the map is a four by 3 square grid). How many routes from X to Y can pat take that have the minimum possible length.

A. 6
B. 8
C. 10
D. 14
E. 16

The correct answer is C. My question: is there a formula i can use for this problem and, if so, can somebody please demonstrate it? I tried the combination formula and it didn't work, despite this problem being categorized an "elementary combination" problem. the book just works out every possible route, one by one, and that just seems a little tedious for the GMAT (who has time to do that on exam day!!??). 0_o

Pat has to walk up 3 times and walk right twice. So, his route is all possible combinations of 3Us and 2Rs. This is akin to creating a 5 letter word with 3 indistinguishable Us and 2 indistinguishable Rs.

Total number of ways = 5!/(3! * 2!) = 20/2 = 10.

(This is because to find the number of distinct permutations of a set of items with indistinguishable items, divide the factorial of the items in the set by the product of number of indistinguishable items.)