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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 22"
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<math>\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160</math>
 
<math>\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160</math>
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==Solution 1==
 
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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 <math>{6 \choose 3}=20</math> ways that Azar could have played. However, two of these ways would lead to Azar winning and must be excluded. This leads to <math>6(20-2)=108</math> ways the board could look after the game is over.
 
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Also, there are two ways for Carl to win with a complete diagonal. For each way, there are six spaces and <math>{6 \choose 3}=20</math> ways Azar could have played. This leads to <math>2\cdot20=40</math> ways the board could look.
 
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In total, there are <math>108+40 = \boxed{\textbf{(D) } 148}</math> ways the board could look after the game is over.
 
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 01:09, 26 November 2021

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 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
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All AMC 12 Problems and Solutions

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