Difference between revisions of "2005 AIME I Problems/Problem 13"
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*'''Case ''' <math>d = 1</math>: It is easy to see only <math>2</math> cases. | *'''Case ''' <math>d = 1</math>: It is easy to see only <math>2</math> cases. | ||
− | *'''Case ''' <math>d = 2</math>: There are two diagonals. We need to generate a string with <math>3</math> <math>R</math>'s, <math>3</math> <math>U</math>'s, and <math>2</math> <math>D</math>'s such that no two <math>R</math>'s or <math>U</math>'s are adjacent. The <math>D</math>'s split the string into three sections (<math>-D-D-</math>): by the [[Pigeonhole principle]] one of the two letters must | + | *'''Case ''' <math>d = 2</math>: There are two diagonals. We need to generate a string with <math>3</math> <math>R</math>'s, <math>3</math> <math>U</math>'s, and <math>2</math> <math>D</math>'s such that no two <math>R</math>'s or <math>U</math>'s are adjacent. The <math>D</math>'s split the string into three sections (<math>-D-D-</math>): by the [[Pigeonhole principle]] all of at least one of the two letters must be all together (i.e., stay in a row). |
− | :If both <math>R</math> and <math>U</math> stay together, then there are <math> | + | :If both <math>R</math> and <math>U</math> stay together, then there are <math>3 \cdot 2=6</math> ways. |
:If either <math>R</math> or <math>U</math> splits, then there are <math>3</math> places to put the letter that splits, which has <math>2</math> possibilities. The remaining letter must divide into <math>2</math> in one section and <math>1</math> in the next, giving <math>2</math> ways. This totals <math>6 + 3\cdot 2\cdot 2 = 18</math> ways. | :If either <math>R</math> or <math>U</math> splits, then there are <math>3</math> places to put the letter that splits, which has <math>2</math> possibilities. The remaining letter must divide into <math>2</math> in one section and <math>1</math> in the next, giving <math>2</math> ways. This totals <math>6 + 3\cdot 2\cdot 2 = 18</math> ways. | ||
*'''Case ''' <math>d = 3</math>: Now <math>2</math> <math>R</math>'s, <math>2</math> <math>U</math>'s, and <math>3</math> <math>D</math>'s, so the string is divided into <math>4</math> partitions (<math>-D-D-D-</math>). | *'''Case ''' <math>d = 3</math>: Now <math>2</math> <math>R</math>'s, <math>2</math> <math>U</math>'s, and <math>3</math> <math>D</math>'s, so the string is divided into <math>4</math> partitions (<math>-D-D-D-</math>). |
Revision as of 20:57, 28 February 2009
Problem
A particle moves in the Cartesian Plane according to the following rules:
- From any lattice point the particle may only move to or
- There are no right angle turns in the particle's path.
How many different paths can the particle take from to ?
Solution
Solution 1
The length of the path (the number of times the particle moves) can range from to ; notice that gives the number of diagonals. Let represent a move to the right, represent a move upwards, and to be a move that is diagonal. Casework upon the number of diagonal moves:
- Case : It is easy to see only cases.
- Case : There are two diagonals. We need to generate a string with 's, 's, and 's such that no two 's or 's are adjacent. The 's split the string into three sections (): by the Pigeonhole principle all of at least one of the two letters must be all together (i.e., stay in a row).
- If both and stay together, then there are ways.
- If either or splits, then there are places to put the letter that splits, which has possibilities. The remaining letter must divide into in one section and in the next, giving ways. This totals ways.
- Case : Now 's, 's, and 's, so the string is divided into partitions ().
- If the 's and 's stay together, then there are places to put them.
- If one of them splits and the other stays together, then there are places to put them, and ways to pick which splits, giving ways.
- If both groups split, then there are ways to arrange them. These add up to ways.
- Case : Now , , 's (). There are places to put , places to put , giving ways.
- Case : It is easy to see only case.
Together, these add up to .
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 possible paths.
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 |