AIME 2010 II · 第 11 题

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 $44+24=\boxed{68}$.

Solution 3

We proceed by complimentary counting.

There are $\binom{9}{5} = 126$ ways to fill the board with five $1$'s and four $0$'s. We subtract the cases where at least two of the eight rows, columns, and long diagonals have all three entries equal. For the sake of simplicity, let a line denote a row/column/diagonal.

Firstly, observe that there can be at most one line of $0$'s and at most two lines of $1$'s. Also, if there is one line of $0$'s, there can at most be one line of $1$'s. Therefore, there are only two "bad" cases we must consider:

Case $1$: There is one line of $0$'s and one line of $1$'s.

In this case, note that both lines must be horizontal or vertical. If both are horizontal, there are $3 \cdot 2 = 6$ ways to select the columns, and $\binom{3}{1} = 3$ ways to place the remaining $0$ in the remaining $3$ grids, yielding a total of $6 \cdot 3 = 18$ bad cases. Similarly, there are $18$ bad cases where both lines are vertical, so $18 + 18 = 36$ cases.

Case $2$: There are two lines of $1$'s.

In this case, note that all five $1$'s are used, and the two lines can be horizontal/vertical, horizontal/diagonal, vertical/diagonal, or diagonal/diagonal, yielding $3 \cdot 3 = 9$, $3 \cdot 2 = 6$, $3 \cdot 2 = 6$, and $1$ case, respectively. $9 + 6 + 6 + 1 = 22$ cases.

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

--OZ30