2021 Fall AMC 12A Problems/Problem 22

Problem

Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a 3-by-3 array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$ . How many ways can the board look after the game is over?

$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$

Solution 1

There are six ways for Carl to win with a complete row or column. For each way, there are six spaces where Carl did not play and Azar could have played. So, there are ${6 \choose 3}=20$ ways that Azar could have played. However, two of these ways would lead to Azar winning and must be excluded. This leads to $6(20-2)=108$ ways the board could look after the game is over.

Also, there are two ways for Carl to win with a complete diagonal. For each way, there are six spaces and ${6 \choose 3}=20$ ways Azar could have played. This leads to $2\cdot20=40$ ways the board could look.

In total, there are $108+40 = \boxed{\textbf{(D) } 148}$ ways the board could look after the game is over.

Solution 2

We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.

$\textbf{Case 1}$ : 3 $O$ are in a horizontal row or a vertical row.

Step 1: We determine the row that 3 $O$ occupy.

The number of ways is 6.

Step 2: We determine the configuration of 3 $X$ .

The number of ways is $\binom{6}{3} - 2 = 18$ .

In this case, following from the rule of product, the number of ways is $6 \cdot 18 = 108$ .

$\textbf{Case 2}$ : 3 $O$ are in a diagonal row.

Step 1: We determine the row that 3 $O$ occupy.

The number of ways is 2.

Step 2: We determine the configuration of 3 $X$ .

The number of ways is $\binom{6}{3} = 20$ .

In this case, following from the rule of product, the number of ways is $2 \cdot 20 = 40$ .

Putting all cases together, the total number of ways is $108 + 40 = 148$ .

Therefore, the answer is $\boxed{\textbf{(D) }148}$ .

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2021 Fall AMC 12A Problems/Problem 22

Problem

Solution 1

Solution 2