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1997 PMWC Problems/Problem I15

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Problem

How many paths from A to B consist of exactly six line segments (vertical, horizontal or inclined)?

size(150);dotfactor=7;pointpen=black;draw(unitsquare);draw((1/3,0)--(1/3,1));draw((2/3,0)--(2/3,1));draw((0,1/3)--(1,1/3));dr...

Solution

Casework:

  • Ignoring the diagonal segments, there are \frac{6!}{3!3!} = 20 paths.
  • Traversing the diagonals, we quickly find that the path must run through exactly 2 diagonals. There are {3\choose2} = 3 pairs of diagonals through which this is possible; quick counting shows us that each pair of diagonals yields 2 paths. So there are 6 more cases here.

In total, we get 20 + 6 = 26 paths.

See also

1997 PMWC (Problems)
Preceded by
Problem I14
Followed by
Problem T1
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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