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

## 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

The T-grid can be consider as a tic-tac-toe board: five $1$'s and four $0$'s.

There are $\dbinom{9}{5} = 126$ ways to fill the board with five $1$'s and four $0$'s. Now we need to subtract the number of bad grids.

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 takes a diagonal, the other cannot win, and if either takes a row/column, all column/row are blocked, so they either both take a row or both take a column.

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

There are $18$ cases.

2. Both takes a column: Using the simlar reasoning, there are $18$ cases.

Case $1$: $36$ cases

Case $2$: Player $1$ wins twice.

1. A row and a column
• $3$ ways to pick the row,
• $3$ to pick the column.
• There are $9$ cases

2. A row/column and a diagonal
• $6$ ways to pick the row/column,
• $2$ to pick the diagonal.
• There are $12$ cases

3. 2 diagonals It is clear that there is only $1$ case.

Case $2$ total: $22$

Thus, the answer is $126-22-36=\boxed{068}$