2010 AIME II Problems/Problem 11

Revision as of 11:58, 6 April 2010 by Azjps (talk | contribs) (See also: minor)


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.


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 $126-22-36=\boxed{068}$

See also

2010 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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