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Difference between revisions of "2001 AIME II Problems/Problem 9"
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== Problem ==
 
== 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 <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>.
  
−
== Solution ==
+
== Solution 1==
−
We can use [[complementary counting]], counting all of the colorings that have at least one red square.
+
We can use [[complementary counting]], counting all of the colorings that have at least one red <math>2\times 2</math> square.
  
−
Counting this directly is messy, so we use [[PIE]]:
+
*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>
 +
*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 be only two ways to pick two squares, and <math>2^2</math> ways to color the other squares. <math>32+8=40</math>
 +
*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.
 +
*For at least four <math>2 \times 2</math> squares, we clearly only have one way.
  
−
For at least one red 2*2 square:
+
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.
  
−
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>
+
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>.
  
−
For at least two squares:
+
==Solution 2==
 +
We consider how many ways we can have 2*2 grid
  
−
There are two cases: those with two red squares on one side and those without red squares on one side.
+
<math>(1)</math>: All the grids are red--<math>1</math> case
  
−
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>
+
<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
  
−
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:
+
<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.
 +
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
  
−
<math>128-40+8-1=95</math>
+
<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>
  
−
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>.
+
<math>(5)</math> Four unit squares are blue, no overcounted case will be considered. there are <math>4\cdot \binom 5 4=20</math>
 +
 
 +
<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.
 +
 
 +
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>
 +
 
 +
~bluesoul
 +
 
 +
==Solution 3 (Case Work)==
 +
 
 +
<cmath>\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}</cmath>
 +
 
 +
Case 1: The 3-by-3 unit-square grid has exactly <math>1</math> 2-by-2 red square
 +
 
 +
Assume the 2-by-2 red square is at <math>C_{11}, C_{12}, C_{21}, C_{22}</math>. To make sure there are no more 2-by-2 red squares, <math>C_{31} \text{and} C_{32}</math> can't both be red and <math>C_{13} \text{and} C_{23}</math> can't both be red. Meaning that there are <math>2^2-1=3</math> coloring methods for <math>C_{31} \text{and} C_{32}</math> and <math>C_{13} \text{and} C_{23}</math>. <math>C_{33}</math> can be colored with either colors. However, the coloring method where <math>C_{23}, C_{32}, C_{33}</math> are all red needs to be removed. For exactly one 2-by-2 red square at <math>C_{11}, C_{12}, C_{21}, C_{22}</math>, there are <math>3 \cdot 3 \cdot 2 -1=17</math> coloring methods. As there are <math>4</math> locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are <math>17 \cdot 4 = 68</math> coloring methods.
 +
 
 +
 
 +
Case 2: The 3-by-3 unit-square grid has exactly <math>2</math> 2-by-2 red squares
 +
 
 +
Case 2.1: <math>2</math> 2-by-2 red squares take up <math>6</math> unit grids
 +
 
 +
Assume the <math>2</math> 2-by-2 red squares are at <math>C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}</math>. To make sure there are no more 2-by-2 red squares, <math>C_{13} \text{and} C_{23}</math> can't both be red. Meaning that there are <math>2^2-1=3</math> coloring methods for <math>C_{13} \text{and} C_{23}</math>. <math>C_{33}</math> can be colored with either colors. However, the coloring method where <math>C_{13}, C_{23}, C_{33}</math> are all red needs to be removed. For exactly <math>2</math> 2-by-2 red squares at <math>C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}</math>, there are <math>3 \cdot 2 -1=5</math> coloring methods. As there are <math>4</math> locations for the <math>2</math> 2-by-2 red squares on the 3-by-3 unit-square grid, there are <math>5 \cdot 4 = 20</math> coloring methods.
 +
 
 +
Case 2.2: <math>2</math> 2-by-2 red squares take up <math>7</math> unit grids
 +
 
 +
Assume the <math>2</math> 2-by-2 red squares are at <math>C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}</math>. To make sure there are no more 2-by-2 red squares, <math>C_{13}</math> and <math>C_{31}</math> can't be red. Meaning that there is <math>1</math> coloring method for <math>C_{13}</math> and <math>C_{31}</math>. For exactly <math>2</math> 2-by-2 red squares at <math>C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}</math>, there is <math>1</math> coloring method. As there are <math>2</math> locations for the <math>2</math> 2-by-2 red squares on the 3-by-3 unit-square grid, there are <math>1 \cdot 2 = 2</math> coloring methods.
 +
 
 +
Hence, if the 3-by-3 unit-square grid has exactly <math>2</math> 2-by-2 red squares, there are <math>20+2 = 22</math> coloring methods.
 +
 
 +
 
 +
Case 3: The 3-by-3 unit-square grid has exactly <math>3</math> 2-by-2 red squares
 +
 
 +
Assume the <math>3</math> 2-by-2 red squares are at <math>C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}</math>. To make sure there are no more 2-by-2 red squares, <math>C_{33}</math> can't be red. Meaning that there is <math>1</math> coloring method for <math>C_{33}</math>. For exactly <math>3</math> 2-by-2 red squares at <math>C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}</math>, there is <math>1</math> coloring method. As there are <math>4</math> locations for the <math>3</math> 2-by-2 red squares on the 3-by-3 unit-square grid, there are <math>1 \cdot 4 = 4</math> coloring methods.
 +
 
 +
 
 +
Case 4: The 3-by-3 unit-square grid has exactly <math>4</math> 2-by-2 red squares
 +
 
 +
If the 3-by-3 unit-square grid has exactly <math>4</math> 2-by-2 red squares, all <math>9</math> unit grids are red and there is <math>1</math> coloring method.
 +
 
 +
 
 +
In total, there are <math>68+22+4+1=95</math> coloring methods with 2-by-2 red squares.
 +
<cmath>\frac{m}{n}=1-\frac{95}{2^9}=\frac{417}{512}</cmath>
 +
 
 +
<cmath>m+n=417+512=\boxed{\textbf{929}}</cmath>
 +
 
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
  
 
== See also ==
 
== See also ==
−
* [[2001 AIME II Problems]]
+
{{AIME box|year=2001|n=II|num-b=8|num-a=10}}
 +
 
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:39, 30 June 2024

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 be 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 grids 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 overcounted 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 (Case Work)

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

Case 1: The 3-by-3 unit-square grid has exactly $1$ 2-by-2 red square

Assume the 2-by-2 red square is at $C_{11}, C_{12}, C_{21}, C_{22}$. To make sure there are no more 2-by-2 red squares, $C_{31} \text{and} C_{32}$ can't both be red and $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{31} \text{and} C_{32}$ and $C_{13} \text{and} C_{23}$. $C_{33}$ can be colored with either colors. However, the coloring method where $C_{23}, C_{32}, C_{33}$ are all red needs to be removed. For exactly one 2-by-2 red square at $C_{11}, C_{12}, C_{21}, C_{22}$, there are $3 \cdot 3 \cdot 2 -1=17$ coloring methods. As there are $4$ locations for the 2-by-2 red square on the 3-by-3 unit-square grid, there are $17 \cdot 4 = 68$ coloring methods.


Case 2: The 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares

Case 2.1: $2$ 2-by-2 red squares take up $6$ unit grids

Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$. To make sure there are no more 2-by-2 red squares, $C_{13} \text{and} C_{23}$ can't both be red. Meaning that there are $2^2-1=3$ coloring methods for $C_{13} \text{and} C_{23}$. $C_{33}$ can be colored with either colors. However, the coloring method where $C_{13}, C_{23}, C_{33}$ are all red needs to be removed. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{31}, C_{32}$, there are $3 \cdot 2 -1=5$ coloring methods. As there are $4$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $5 \cdot 4 = 20$ coloring methods.

Case 2.2: $2$ 2-by-2 red squares take up $7$ unit grids

Assume the $2$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$. To make sure there are no more 2-by-2 red squares, $C_{13}$ and $C_{31}$ can't be red. Meaning that there is $1$ coloring method for $C_{13}$ and $C_{31}$. For exactly $2$ 2-by-2 red squares at $C_{11}, C_{12}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$, there is $1$ coloring method. As there are $2$ locations for the $2$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 2 = 2$ coloring methods.

Hence, if the 3-by-3 unit-square grid has exactly $2$ 2-by-2 red squares, there are $20+2 = 22$ coloring methods.


Case 3: The 3-by-3 unit-square grid has exactly $3$ 2-by-2 red squares

Assume the $3$ 2-by-2 red squares are at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$. To make sure there are no more 2-by-2 red squares, $C_{33}$ can't be red. Meaning that there is $1$ coloring method for $C_{33}$. For exactly $3$ 2-by-2 red squares at $C_{11}, C_{12}, C_{13}, C_{21}, C_{22}, C_{23}, C_{32}, C_{33}$, there is $1$ coloring method. As there are $4$ locations for the $3$ 2-by-2 red squares on the 3-by-3 unit-square grid, there are $1 \cdot 4 = 4$ coloring methods.


Case 4: The 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares

If the 3-by-3 unit-square grid has exactly $4$ 2-by-2 red squares, all $9$ unit grids are red and there is $1$ coloring method.


In total, there are $68+22+4+1=95$ coloring methods with 2-by-2 red squares. \[\frac{m}{n}=1-\frac{95}{2^9}=\frac{417}{512}\]

\[m+n=417+512=\boxed{\textbf{929}}\]

~isabelchen

See also

2001 AIME II (Problems • Answer Key • Resources)
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