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

(Problem)
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== 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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Counting this directly is messy, so we use [[PIE]]:
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For at least one red 2*2 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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For at least two squares:
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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 alternatively add and subtract:
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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>.
  
 
== See also ==
 
== See also ==
 
* [[2001 AIME II Problems]]
 
* [[2001 AIME II Problems]]

Revision as of 14:23, 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 square.

Counting this directly is messy, so we use PIE:

For at least one red 2*2 square:

There are four 2*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 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. $32+8=40$

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 alternatively add and subtract:

$128-40+8-1=95$

There are $2^9=512$ ways to paint the 3*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-by-2 red square is $\frac{417}{512}$, and $417+512=\boxed{929}$.

See also

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