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

Case 1: Carl wins with a complete row or column.

Carl can win in three rows or three columns.

For each row and column, there are six spaces where Carl did not play (so Azar could have played) and {6 \choose 3} ways Azar could play. However, two of these ways would lead to Azar winning, so they must be excluded.

This leads to $6({6 \choose 3}-2)=108$ possible board conditions.

Case 2: Carl wins with a diagonal.

Carl can win in two diagonals.

For each diagonal, there are six spaces where Carl did not play (so Azar could have played) and {6 \choose 3} ways Azar could play.

This leads to $2{6 \choose 3}=40$ possible board conditions.

Therefore, there are a total of $108+40 \implies \boxed{\textbf{(D) } 148}$ board conditions.

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 20
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2021 Fall AMC 12A Problems/Problem 22

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