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#### find_the_fun

##### Active member

- Feb 1, 2012

- 166

The answer key gives 14x13 as the solution to part c. I understand there are 14 ways to get to point C but depending on which road can no longer be used, there is anywhere between 10 and 13 ways to return home. For example, say R8 was taken to get to C. Then there are 14-1 ways to make the return trip (just don't use R8). However, if R1R5 is used then we can't use either R1 (so now there are 3x3+2=11 ways to make the trip) or R5 (so now there are 2x4+2=10 ways to make the trip). What is wrong with my reasoning?