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Difference between revisions of "2001 AIME II Problems/Problem 9"
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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>.
 
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>.
 +
 +
==solution 2==
 +
We consider how many ways we can have 2*2 grid
 +
 +
<math>(1)</math>: All the girds are red--<math>1</math> case
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<math>(2)</math>: One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid <math>9-1=8</math> cases
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<math>(3)</math>: Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, <math>8</math> cases.
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Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. <math>\binom 9 2-8-2=26</math> cases
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<math>(4)</math> Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are <math>4\cdot (\binom 5 3)-4=36</math>
 +
<math>(5)</math> Four unit squares are blue, no overcomunted case will be considered. there are <math>4\cdot \binom 5 4=20</math>
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<math>(6)</math> Five unit squares are blue, <math>4</math> cases in all
 +
 +
Sum up those cases, there are <math>1+8+26+36+20+4=95</math> cases that a 2*2 grid can be formed.
 +
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In all, there are <math>2^9=512</math> possible ways to paint the big square, so the answer is <math>1-\frac{95}{512}=\frac{417}{512}</math> leads to <math>\boxed{929}</math>
  
 
== See also ==
 
== See also ==

Revision as of 12:54, 15 May 2022

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 $417+512=\boxed{929}$.

solution 2

We consider how many ways we can have 2*2 grid

$(1)$: All the girds are red--$1$ case $(2)$: One unit square is blue--The blue lies on the center of the bigger square, makes no 2*2 grid $9-1=8$ cases $(3)$: Two unit squares are blue--one of the squares lies in the center of the bigger square, makes no 2*2 grid, $8$ cases. Or, two squares lie on second column, first row, second column third row; second row first column, second row third column, 2 extra cases. $\binom 9 2-8-2=26$ cases $(4)$ Three unit squares are blue. We find that if a 2*2 square is formed, there are 5 extra unit squares can be painted. But cases that three squares in the same column or same row is overcomunted. So in this case, there are $4\cdot (\binom 5 3)-4=36$ $(5)$ Four unit squares are blue, no overcomunted case will be considered. there are $4\cdot \binom 5 4=20$ $(6)$ Five unit squares are blue, $4$ cases in all

Sum up those cases, there are $1+8+26+36+20+4=95$ cases that a 2*2 grid can be formed.

In all, there are $2^9=512$ possible ways to paint the big square, so the answer is $1-\frac{95}{512}=\frac{417}{512}$ leads to $\boxed{929}$

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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