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Difference between revisions of "1997 PMWC Problems/Problem I15"

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== Problem ==
 
== Problem ==
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How many paths from A to B consist of exactly six line
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How many paths from <math>A</math> to <math>B</math> consist of exactly six line segments (vertical, horizontal or inclined)?
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segments (vertical, horizontal or inclined)?
 
  
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<asy>size(150);
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<asy>
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dotfactor=7;
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for(int i = 0; i < 3; ++i){
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pointpen=black;
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draw((0,i+1)--(0,i)--(4,i)--(4,i+1));
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draw(unitsquare);
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draw((4/3,i+1)--(4/3,i)--(8/3,i+1)--(8/3,i));
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draw((1/3,0)--(1/3,1));draw((2/3,0)--(2/3,1));
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}
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draw((0,1/3)--(1,1/3));draw((0,2/3)--(1,2/3));
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draw((0,3)--(4,3));
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draw((1/3,0)--(2/3,1/3));draw((1/3,1/3)--(2/3,2/3));draw((1/3,2/3)--(2/3,1));
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label("$A$",(0,0),SW);
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for(int i=0;i<4;++i)
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label("$B$",(4,3),NE);
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  for(int j=0;j<4;++j)
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//Credit to chezbgone2 for the diagram</asy>
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    dot((i/3,j/3));
 
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MP("\mathrm{A}",D((0,0)),SW,fontsize(9));
 
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MP("\mathrm{B}",D((1,1)),NE,fontsize(9));</asy>
 
  
 
== Solution ==
 
== Solution ==
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In total, we get <math>20 + 6 = 26</math> paths.
 
In total, we get <math>20 + 6 = 26</math> paths.
  
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== See also ==
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== 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 $A$ to $B$ consist of exactly six line segments (vertical, horizontal or inclined)?

[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((0,3)--(4,3)); label("$A$",(0,0),SW); label("$B$",(4,3),NE); //Credit to chezbgone2 for the diagram[/asy]

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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