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Difference between revisions of "2010 AIME II Problems/Problem 11"

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== Solution ==
 
== Solution ==
  
−
The <i>T-grid</i> can be consider as a tic-tac-toe board: five <math>1</math>'s and four <math>0</math>'s.  
+
=== Solution 1 ===
 +
The <i>T-grid</i> can be considered as a tic-tac-toe board: five <math>1</math>'s (or X's) and four <math>0</math>'s (or O's).  
  
−
There are <math>\dbinom{9}{5} = 126</math>  ways to fill the board with five <math>1</math>'s and four <math>0</math>'s. Now we need to subtract the number of bad grids.
+
There are only <math>\dbinom{9}{5} = 126</math>  ways to fill the board with five <math>1</math>'s and four <math>0</math>'s. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone has won twice.
  
 
Let three-in-a-row/column/diagonal be a "win" and let player <math>0</math> be the one that fills in <math>0</math> and player <math>1</math> fills in <math>1</math>.
 
Let three-in-a-row/column/diagonal be a "win" and let player <math>0</math> be the one that fills in <math>0</math> and player <math>1</math> fills in <math>1</math>.
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<b>Case <math>1</math>:</b> Each player wins once.  
 
<b>Case <math>1</math>:</b> Each player wins once.  
  
−
If player takes a diagonal, the other cannot win, and
+
If player X takes a diagonal, player Y cannot win. If either takes a row, all the columns are blocked, and visa versa. Therefore, they either both take a row or they both take a column.
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if either takes a row/column, all column/row are blocked, so they either both take a row or both take a column.
 
  
 
<OL>
 
<OL>
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<LI>Both takes a row:  
+
<LI>Both take a row:  
  
 
<UL>
 
<UL>
−
<LI><math>3</math> ways for player <math>1</math> to pick a row, </LI>
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<LI>There are <math>3</math> ways for player <math>1</math> to pick a row, </LI>
  
−
<LI><math>2</math> ways for player <math>0</math>, </LI>
+
<LI><math>2</math> ways for player <math>0</math> to pick a row and, </LI>
  
 
<LI><math>3</math> ways for player <math>0</math> to take a single box in the remaining row. </LI>
 
<LI><math>3</math> ways for player <math>0</math> to take a single box in the remaining row. </LI>
 
</UL>
 
</UL>
  
−
There are <math>18</math> cases.
+
Therefore, there are <math>3 \cdot 2 \cdot 3=18</math> cases.
 
</LI>
 
</LI>
  
−
<LI>Both takes a column:  
+
<LI>Both take a column:  
  
−
Using the simlar reasoning, there are <math>18</math> cases.
+
Using similar reasoning, there are <math>18</math> cases.
 
</OL>
 
</OL>
  
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<br/>
 
<br/>
  
−
<b>Case <math>2</math>:</b> Player <math>1</math> wins twice.  
+
<b>Case <math>2</math>:</b> Player <math>1</math> wins twice. (Player <math>0</math> cannot win twice because he only has 4 moves.)
  
 
<OL>
 
<OL>
−
<LI>A row and a column
+
<LI>Player <math>1</math> picks a row and a column.
 
<UL>
 
<UL>
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<LI><math>3</math> ways to pick the row, </LI>
+
<LI>There are <math>3</math> ways to pick the row</LI>
−
<LI><math>3</math> to pick the column. </LI>
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<LI>and <math>3</math> ways to pick the column. </LI>
−
There are <math>9</math> cases
+
So there are <math>9</math> cases.
 
</UL></LI>
 
</UL></LI>
  
−
<LI>A row/column and a diagonal
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<LI>Player <math>1</math> picks a row/column and a diagonal.
 
<UL>
 
<UL>
−
<LI><math>6</math> ways to pick the row/column, </LI>
+
<LI>There are <math>6</math> ways to pick the row/column </LI>
−
<LI><math>2</math> to pick the diagonal. </LI>
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<LI>and <math>2</math> to pick the diagonal, </LI>
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There are <math>12</math> cases
+
</UL>
−
</UL></LI>
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so there are <math>12</math> cases
 +
</LI>
  
 
<LI>2 diagonals
 
<LI>2 diagonals
−
 
+
<UL>
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It is clear that there is only <math>1</math> case.</LI>
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<LI>It is clear that there is only <math>1</math> case. (Make a X).</LI>
 +
</UL>
 +
</LI>
 
</OL>
 
</OL>
  
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<br/>
 
<br/>
  
−
Thus, the answer is <math>126-22-36=\boxed{068}</math>
+
Thus, the answer is <math>126-22-36=\boxed{68}</math>
 +
 
 +
=== Solution 2 ===
 +
 
 +
We can use [[generating functions]] to compute the case that no row or column is completely filled with <math>1</math>'s. Given a row, let <math>a</math>, <math>b</math>, <math>c</math> be the events that the first, second, third respective squares are <math>1</math>'s. Then the generating function representing the possible events that exclude a row of <math>1,1,1</math> or <math>0,0,0</math> from occuring is
 +
<cmath>ab+bc+ca+a+b+c.</cmath>
 +
Therefore, the generating function representing the possible grids where no row is filled with <math>0</math>'s and <math>1</math>'s is
 +
<cmath>P(a,b,c)=((ab+bc+ca)+(a+b+c))^3.</cmath>
 +
We expand this by the Binomial Theorem to find
 +
<cmath>P(a,b,c)=(ab+bc+ca)^3+3(ab+bc+ca)^2(a+b+c)+3(ab+bc+ca)(a+b+c)^2+(a+b+c)^3.</cmath>
 +
Recall that our grid has five <math>1</math>'s, hence we only want terms where the sum of the exponents is <math>5</math>. This is given by
 +
<cmath>3(ab+bc+ca)^2(a+b+c).</cmath>
 +
When we expand this, we find
 +
<cmath>3(2abc(a+b+c)+a^2b^2+b^2c^2+c^2a^2)(a+b+c).</cmath>
 +
We also want to make sure that each of <math>a</math>, <math>b</math>, <math>c</math> appears at least once (so there is no column filled with <math>0</math>'s) and the power of each of <math>a</math>, <math>b</math>, <math>c</math> is not greater than or equal to <math>3</math> (so there is no column filled with <math>1</math>'s). The sum of the coefficients of the above polynomial is clearly <math>81</math> (using <math>a,b,c=1</math>), and the sum of the coefficients of the terms <math>a^3bc</math>, <math>ab^3c</math>, and <math>abc^3</math> is <math>6+6+6+3+3+3+3+3+3=36</math>, hence the sum of the coefficients of the desired terms is <math>81-36=45</math>. This counts the number of grids where no column or row is filled with <math>0</math>'s or <math>1</math>'s. However, we could potentially have both diagonals filled with <math>1</math>'s, but this is the only exception to our <math>45</math> possibilities, hence the number of <math>T</math>-grids with no row or column filled with the same digit is <math>44</math>.
 +
 
 +
On the other hand, if a row (column) is filled with <math>0</math>'s, then by the Pigeonhole Principle, another row (column) must be filled with <math>1</math>'s. Hence this is impossible, so all other possible <math>T</math>-grids have a row (column) filled with <math>1</math>'s. If the top row is filled with <math>1</math>'s, then we have two <math>1</math>'s left to place. Clearly they cannot go in the same row, because then the other row is filled with <math>0</math>'s. They also cannot appear in the same column. This leaves <math>3\cdot 2</math> arrangements--3 choices for the location of the <math>1</math> in the second row, and 2 choices for the location of the <math>1</math> in the last row. However, two of these arrangements will fill a diagonal with <math>1</math>'s. Hence there are only <math>4</math> <math>T</math>-grids where the top row is filled with <math>1</math>'s. The same argument applies if any other row or column is filled with <math>1</math>'s. Hence there are <math>4\cdot 6=24</math> such <math>T</math>-grids.
 +
 
 +
Thus the answer is <math>44+24=\boxed{68}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:58, 8 January 2023

Problem

Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:

  1. Exactly five of the entries are $1$'s, and the remaining four entries are $0$'s.
  2. Among the eight rows, columns, and long diagonals (the long diagonals are $\{a_{13},a_{22},a_{31}\}$ and $\{a_{11},a_{22},a_{33}\}$, no more than one of the eight has all three entries equal.

Find the number of distinct T-grids.


Solution

Solution 1

The T-grid can be considered as a tic-tac-toe board: five $1$'s (or X's) and four $0$'s (or O's).

There are only $\dbinom{9}{5} = 126$ ways to fill the board with five $1$'s and four $0$'s. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone has won twice.

Let three-in-a-row/column/diagonal be a "win" and let player $0$ be the one that fills in $0$ and player $1$ fills in $1$.

Case $1$: Each player wins once.

If player X takes a diagonal, player Y cannot win. If either takes a row, all the columns are blocked, and visa versa. Therefore, they either both take a row or they both take a column.

  1. Both take a row:
    • There are $3$ ways for player $1$ to pick a row,
    • $2$ ways for player $0$ to pick a row and,
    • $3$ ways for player $0$ to take a single box in the remaining row.

    Therefore, there are $3 \cdot 2 \cdot 3=18$ cases.

  2. Both take a column: Using similar reasoning, there are $18$ cases.

Case $1$: $36$ cases


Case $2$: Player $1$ wins twice. (Player $0$ cannot win twice because he only has 4 moves.)

  1. Player $1$ picks a row and a column.
    • There are $3$ ways to pick the row
    • and $3$ ways to pick the column.

      • So there are $9$ cases.

  2. Player $1$ picks a row/column and a diagonal.
    • There are $6$ ways to pick the row/column
    • and $2$ to pick the diagonal,

    so there are $12$ cases

  3. 2 diagonals
    • It is clear that there is only $1$ case. (Make a X).

Case $2$ total: $22$


Thus, the answer is $126-22-36=\boxed{68}$

Solution 2

We can use generating functions to compute the case that no row or column is completely filled with $1$'s. Given a row, let $a$, $b$, $c$ be the events that the first, second, third respective squares are $1$'s. Then the generating function representing the possible events that exclude a row of $1,1,1$ or $0,0,0$ from occuring is \[ab+bc+ca+a+b+c.\] Therefore, the generating function representing the possible grids where no row is filled with $0$'s and $1$'s is \[P(a,b,c)=((ab+bc+ca)+(a+b+c))^3.\] We expand this by the Binomial Theorem to find \[P(a,b,c)=(ab+bc+ca)^3+3(ab+bc+ca)^2(a+b+c)+3(ab+bc+ca)(a+b+c)^2+(a+b+c)^3.\] Recall that our grid has five $1$'s, hence we only want terms where the sum of the exponents is $5$. This is given by \[3(ab+bc+ca)^2(a+b+c).\] When we expand this, we find \[3(2abc(a+b+c)+a^2b^2+b^2c^2+c^2a^2)(a+b+c).\] We also want to make sure that each of $a$, $b$, $c$ appears at least once (so there is no column filled with $0$'s) and the power of each of $a$, $b$, $c$ is not greater than or equal to $3$ (so there is no column filled with $1$'s). The sum of the coefficients of the above polynomial is clearly $81$ (using $a,b,c=1$), and the sum of the coefficients of the terms $a^3bc$, $ab^3c$, and $abc^3$ is $6+6+6+3+3+3+3+3+3=36$, hence the sum of the coefficients of the desired terms is $81-36=45$. This counts the number of grids where no column or row is filled with $0$'s or $1$'s. However, we could potentially have both diagonals filled with $1$'s, but this is the only exception to our $45$ possibilities, hence the number of $T$-grids with no row or column filled with the same digit is $44$.

On the other hand, if a row (column) is filled with $0$'s, then by the Pigeonhole Principle, another row (column) must be filled with $1$'s. Hence this is impossible, so all other possible $T$-grids have a row (column) filled with $1$'s. If the top row is filled with $1$'s, then we have two $1$'s left to place. Clearly they cannot go in the same row, because then the other row is filled with $0$'s. They also cannot appear in the same column. This leaves $3\cdot 2$ arrangements--3 choices for the location of the $1$ in the second row, and 2 choices for the location of the $1$ in the last row. However, two of these arrangements will fill a diagonal with $1$'s. Hence there are only $4$ $T$-grids where the top row is filled with $1$'s. The same argument applies if any other row or column is filled with $1$'s. Hence there are $4\cdot 6=24$ such $T$-grids.

Thus the answer is $44+24=\boxed{68}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
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Problem 10 		Followed by
Problem 12
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