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

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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>.
 
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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== Solution ==
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== Solution 1==
 
We can use [[complementary counting]], counting all of the colorings that have at least one red <math>2\times 2</math> square.
 
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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~bluesoul
 
~bluesoul
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 +
==Solution 3==
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 +
<cmath>\begin{array}{|c|c|c|}
 +
\hline
 +
C_{11} & C_{12} & C_{13}\\
 +
\hline
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C_{21} & C_{22} & C_{23}\\
 +
\hline
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C_{31} & C_{32} & C_{33}\\
 +
\hline
 +
\end{array}</cmath>
 +
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
== See also ==
 
== See also ==

Revision as of 11:31, 12 April 2023

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 1

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 $\boxed{929}$

~bluesoul

Solution 3

\[\begin{array}{|c|c|c|} \hline C_{11} & C_{12} & C_{13}\\ \hline C_{21} & C_{22} & C_{23}\\ \hline C_{31} & C_{32} & C_{33}\\ \hline \end{array}\]


~isabelchen

See also

2001 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AIME Problems and Solutions

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