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Difference between revisions of "2001 AIME II Problems/Problem 9"

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== Problem ==
 
== Problem ==
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Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
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Each unit [[square]] of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The [[probability]] of obtaining a grid that does not have a 2-by-2 red square is <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.
  
 
== Solution ==
 
== Solution ==
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We can use [[complementary counting]], counting all of the colorings that have at least one red square.
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We can use [[complementary counting]], counting all of the colorings that have at least one red <math>2\times 2</math> square.
  
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Counting this directly is messy, so we use [[PIE]]:
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*For at least one red <math>2 \times 2</math> square:
 +
:There are four <math>2 \times 2</math> squares to choose which one will be red. Then there are <math>2^5</math> ways to color the rest of the squares. <math>4*32=128</math>
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*For at least two <math>2 \times 2</math> squares:
 +
:There are two cases: those with two red squares on one side and those without red squares on one side.
 +
:The first case is easy: 4 ways to choose which the side the squares will be on, and <math>2^3</math> ways to color the rest of the squares, so 32 ways to do that. For the second case, there will by only two ways to pick two squares, and <math>2^2</math> ways to color the other squares. <math>32+8=40</math>
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*For at least three <math>2 \times 2</math> squares:
 +
:Choosing three such squares leaves only one square left, with four places to place it. This is <math>2 \cdot 4 = 8</math> ways.
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*For at least four <math>2 \times 2</math> squares, we clearly only have one way.
  
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For at least one red 2*2 square:
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By the [[Principle of Inclusion-Exclusion]], there are (alternatively subtracting and adding) <math>128-40+8-1=95</math> ways to have at least one red <math>2 \times 2</math> square.
  
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There are four 2*2 squares to choose which one will be red. Then there are <math>2^5</math> ways to color the rest of the squares. <math>4*32=128</math>
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There are <math>2^9=512</math> ways to paint the <math>3 \times 3</math> square with no restrictions, so there are <math>512-95=417</math> ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a <math>2 \times 2</math> red square is <math>\frac{417}{512}</math>, and <math>417+512=\boxed{929}</math>.
  
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For at least two squares:
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== See also ==
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{{AIME box|year=2001|n=II|num-b=8|num-a=10}}
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There are two cases: those with two red squares on one side and those without red squares on one side.
 
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The first case is easy: 4 ways to choose which the side the squares will be on, and <math>2^3</math> ways to color the rest of the squares, so 32 ways to do that. For the second case, there will by only two ways to pick two squares, and <math>2^2</math> ways to color the other squares. <math>32+8=40</math>
 
  
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We proceed with similar ways with three and four red squares, to get that there are 8 ways to get three red squares, and one way to get four red squares. We then alternate adding and subtracting:
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[[Category:Intermediate Combinatorics Problems]]
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<math>128-40+8-1=95</math>
 
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There are <math>2^9=512</math> ways to paint the 3*3 square with no restrictions, so there are <math>512-95=417</math> ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a 2-by-2 red square is <math>\frac{417}{512}</math>, and <math>417+512=\boxed{929}</math>.
 
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== See also ==
 
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* [[2001 AIME II Problems]]
 

Revision as of 15:51, 30 October 2007

Problem

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

We can use complementary counting, counting all of the colorings that have at least one red $2\times 2$ square.

  • For at least one red $2 \times 2$ square:
There are four $2 \times 2$ squares to choose which one will be red. Then there are $2^5$ ways to color the rest of the squares. $4*32=128$
  • For at least two $2 \times 2$ squares:
There are two cases: those with two red squares on one side and those without red squares on one side.
The first case is easy: 4 ways to choose which the side the squares will be on, and $2^3$ ways to color the rest of the squares, so 32 ways to do that. For the second case, there will by only two ways to pick two squares, and $2^2$ ways to color the other squares. $32+8=40$
  • For at least three $2 \times 2$ squares:
Choosing three such squares leaves only one square left, with four places to place it. This is $2 \cdot 4 = 8$ ways.
  • For at least four $2 \times 2$ squares, we clearly only have one way.

By the Principle of Inclusion-Exclusion, there are (alternatively subtracting and adding) $128-40+8-1=95$ ways to have at least one red $2 \times 2$ square.

There are $2^9=512$ ways to paint the $3 \times 3$ square with no restrictions, so there are $512-95=417$ ways to paint the square with the restriction. Therefore, the probability of obtaining a grid that does not have a $2 \times 2$ red square is $\frac{417}{512}$, and $417+512=\boxed{929}$.

See also

2001 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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