The shortest way to go from A to X ( Chippewa square of X town) is the straight line ( or the diagonal of the big rectangle) from A to X. However, there's no such a way. So we must go vertically then horizontally in some way to get to X. The shortest way can only be a distant of 3 horizontal periods & 2 vertical periods.

So the no. of way to go from A to X is a permutation, in which H is horizontal periods, V is vertical periods.
If we go 2 horizontal periods first then go a 3 vertical periods to get to X, then that way can be expressed as : HHVVV

Total No. of ways : 5P5/(3!*2!*) = 5!/(3!*2!) = 10

mj78ind wrote:Marty wants to go to the Chippewa Square of Town X (refer attached file). He has to start form point A and all route segments are equidistant, how many shortest paths can Marty take?
1. 7
2. 8
3. 10
4. 12
5. 18

The OA is C

What I want to understand is if there is a robust way of solving such questions, rather than just counting the possibilities? Say if we change the number of rightward segments to 3 and upward segments to 4 ........

Thanks!

i counted it and its C the answer
the ways can be ( i suppose it left and go ahead =T)
LLTTT
LTLTT
LTTLT
LTTTL
TLLTT
TLTLT
TLTTL
TTLLT
TTLTL
TTTL
thus 10 ways