Difference between revisions of "1997 PMWC Problems/Problem I15"
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== Problem == | == Problem == | ||
− | How many paths from A to B consist of exactly six line | + | How many paths from <math>A</math> to <math>B</math> consist of exactly six line segments (vertical, horizontal or inclined)? |
− | segments (vertical, horizontal or inclined)? | ||
− | <asy> | + | <asy> |
− | + | for(int i = 0; i < 3; ++i){ | |
− | + | draw((0,i+1)--(0,i)--(4,i)--(4,i+1)); | |
− | + | draw((4/3,i+1)--(4/3,i)--(8/3,i+1)--(8/3,i)); | |
− | draw(( | + | } |
− | + | draw((0,3)--(4,3)); | |
− | draw(( | + | label("$A$",(0,0),SW); |
− | + | label("$B$",(4,3),NE); | |
− | + | //Credit to chezbgone2 for the diagram</asy> | |
− | |||
− | |||
− | |||
== Solution == | == Solution == | ||
Line 24: | Line 20: | ||
In total, we get <math>20 + 6 = 26</math> paths. | In total, we get <math>20 + 6 = 26</math> paths. | ||
− | == See | + | == See Also == |
{{PMWC box|year=1997|num-b=I14|num-a=T1}} | {{PMWC box|year=1997|num-b=I14|num-a=T1}} | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] |
Latest revision as of 14:37, 20 April 2014
Problem
How many paths from to
consist of exactly six line segments (vertical, horizontal or inclined)?
Solution
- Ignoring the diagonal segments, there are
paths.
- Traversing the diagonals, we quickly find that the path must run through exactly 2 diagonals. There are
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 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 |