Difference between revisions of "2005 AIME I Problems/Problem 13"

Revision as of 16:00, 6 January 2008

Problem

A particle moves in the Cartesian Plane according to the following rules:

From any lattice point $(a,b),$ the particle may only move to $(a+1,b), (a,b+1),$ or $(a+1,b+1).$
There are no right angle turns in the particle's path.

How many different paths can the particle take from $(0,0)$ to $(5,5)$ ?

Solution

Solution 1

The length of the path (the number of times the particle moves) can range from $l = 5$ to $9$ ; notice that $d = 10-l$ gives the number of diagonals. Let $R$ represent a move to the right, $U$ represent a move upwards, and $D$ to be a move that is diagonal. Casework upon the number of diagonal moves:

Case $d = 1$ : It is easy to see only $2$ cases.
Case $d = 2$ : There are two diagonals. We need to generate a string with $3$ $R$ 's, $3$ $U$ 's, and $2$ $D$ 's such that no two $R$ 's or $U$ 's are adjacent. The $D$ 's split the string into three sections ( $-D-D-$ ): by the Pigeonhole principle one of the two letters must remain together then (eg, stay in a row).

If both $R$ and $U$ stay together, then there are ${3\choose 2}=6$ ways.

If either $R$ or $U$ splits, then there are $3$ places to put the letter that splits, which has $2$ possibilities. The remaining letter must divide into $2$ in one section and $1$ in the next, giving $2$ ways. This totals $6 + 3\cdot 2\cdot 2 = 18$ ways.

Case $d = 3$ : Now $2$ $R$ 's, $2$ $U$ 's, and $3$ $D$ 's, so the string is divided into $4$ partitions ( $-D-D-D-$ ).

If the $R$ 's and $U$ 's stay together, then there are $4 \cdot 3 = 12$ places to put them.

If one of them splits and the other stays together, then there are $4 \cdot {3\choose 2}$ places to put them, and $2$ ways to pick which splits, giving $4 \cdot 3 \cdot 2 = 24$ ways.

If both groups split, then there are ${4\choose 2}=6$ ways to arrange them. These add up to $12 + 24 + 6 = 42$ ways.

Case $d = 4$ : Now $1$ $R$ , $1$ $U$ , $4$ $D$ 's ( $-D-D-D-D-$ ). There are $5$ places to put $R$ , $4$ places to put $U$ , giving $20$ ways.
Case $d = 5$ : It is easy to see only $1$ case.

Together, these add up to $2 + 18 + 42 + 20 + 1 = \boxed{83}$ .

Solution 2

Another possibility is to use block-walking and recursion: for each vertex, the number of ways to reach it is the number of ways to reach the vertex to its left not coming from down plus the number of ways to reach the vertex below it not coming from the left plus the number of ways to reach the vertex diagonally down and to the left from any direction. As a result, we find 28 ways to reach (5, 5) coming from below, 28 ways to reach it coming from the left and 27 ways to reach it coming diagonally for a total of $083$ possible paths.

Template:Incomplete

@@ Line 28: / Line 28: @@
 Another possibility is to use block-walking and [[recursion]]: for each vertex, the number of ways to reach it is the number of ways to reach the vertex to its left not coming from down plus the number of ways to reach the vertex below it not coming from the left plus the number of ways to reach the vertex diagonally down and to the left from any direction.  As a result, we find 28 ways to reach (5, 5) coming from below, 28 ways to reach it coming from the left and 27 ways to reach it coming diagonally for a total of <math>083</math> possible paths.
-{{incomplete}}
+{{incomplete|solution}}
 == See also ==

2005 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search