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2012 AMC 10B Problem 25

Problem 25 of 25HarderCounting & Probability

A bug travels from AA to BB along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?

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Solution

Classify a path by the set SS of backward arrows it uses. If S=S=\varnothing, the path is determined by choosing one forward arrow in each column, giving 2244422=2102\cdot2\cdot4\cdot4\cdot4\cdot2\cdot2=2^{10} paths. If SS uses only the left backward arrow, there are 282^8 paths, and by symmetry the same for only the right backward arrow. If it uses both outer backward arrows but not the middle one, there are 262^6 paths. If SS uses only the middle backward arrow, there are 292^9 paths. If it uses the middle arrow and exactly one outer backward arrow, there are 272^7 paths for each choice of outer arrow. If it uses all three backward arrows, there are 252^5 paths. The total is 2102^{10} +228+2\cdot2^8 +26+2^6 +29+2^9 +227+2\cdot2^7 +25=2400+2^5=2400. Thus, E is the correct answer.

More practice

Concepts: lattice paths · casework · multiplication principle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.