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

Revision as of 18:37, 24 June 2018

Problem

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

Exactly five of the entries are $1$ 's, and the remaining four entries are $0$ 's.
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.

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.
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.)

Player picks a row and a column.
- There are $3$ ways to pick the row
- and $3$ ways to pick the column.
Player 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
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 $44+24=\boxed{68}$ .

@@ Line 33: / Line 33: @@
 </UL>
-Therefore, there are <math>3 \cdot 2 \cdot 2=18</math> cases.
+Therefore, there are <math>3 \cdot 2 \cdot 3=18</math> cases.
 </LI>

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