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

Revision as of 14:04, 16 April 2008

Problem

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

$[asy]size(150); dotfactor=7; pointpen=blue; draw(unitsquare); draw((1/3,0)--(1/3,1));draw((2/3,0)--(2/3,1)); draw((0,1/3)--(1,1/3));draw((0,2/3)--(1,2/3)); draw((1/3,0)--(2/3,1/3));draw((1/3,1/3)--(2/3,2/3));draw((1/3,2/3)--(2/3,1)); for(int i=0;i<4;++i) for(int j=0;j<4;++j) dot((i/3,j/3)); MP("\mathrm{A}",D((0,0)),SW,fontsize(9)+blue); MP("\mathrm{B}",D((1,1)),NE,fontsize(9)+blue);[/asy]$

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.

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

Revision as of 14:04, 16 April 2008

Problem

Solution

See also

@@ Line 3: / Line 3: @@
 segments (vertical, horizontal or inclined)?
-[[Image:1997_PMWC-I15.png]]
+<asy>size(150);
+dotfactor=7;
+pointpen=blue;
+draw(unitsquare);
+draw((1/3,0)--(1/3,1));draw((2/3,0)--(2/3,1));
+draw((0,1/3)--(1,1/3));draw((0,2/3)--(1,2/3));
+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));
+for(int i=0;i<4;++i)
+  for(int j=0;j<4;++j)
+    dot((i/3,j/3));
+MP("\mathrm{A}",D((0,0)),SW,fontsize(9)+blue);
+MP("\mathrm{B}",D((1,1)),NE,fontsize(9)+blue);</asy>
 == Solution ==