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2021 Fall AMC 12A Problems/Problem 22: Difference between revisions

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==Problem==
==Problem==
Azar and Carl play a game of tic-tac-toe. Azar places an in <math>X</math> one of the boxes in a <math>3</math>-by-<math>3</math> array of boxes, then Carl places an <math>O</math> in one of the remaining boxes. After that, Azar places an <math>X</math> 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 <math>O</math>. How many ways can the board look after the game is over?
Azar and Carl play a game of tic-tac-toe. Azar places an <math>X</math> one of the boxes in a <math>3</math>-by-<math>3</math> array of boxes, then Carl places an <math>O</math> in one of the remaining boxes. After that, Azar places an <math>X</math> in one of the remaining boxes, and so on until all 9 boxes are filled or one of the players has 3 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 <math>O</math>. How many ways can the board look after the game is over?


<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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Step 2: We determine the configuration of 3 <math>X</math>.
Step 2: We determine the configuration of 3 <math>X</math>.


The number of ways is <math>\binom{6}{3} - 2 = 18</math>.
The number of ways is <math>\binom{6}{3} - 2 = 18</math>. (6 open spots to put the <math>X</math> and subtracting cases where they fill row/column)


In this case, following from the rule of product, the number of ways is <math>6 \cdot 18 = 108</math>.
In this case, following from the rule of product, the number of ways is <math>6 \cdot 18 = 108</math>.
Line 36: Line 36:


~Steven Chen (www.professorchenedu.com)
~Steven Chen (www.professorchenedu.com)
== Video Solution by OmegaLearn ==
https://youtu.be/kxgUdv_L-ys?t=796
~ pi_is_3.14
==Video Solution by Mathematical Dexterity==
https://www.youtube.com/watch?v=OpRk-iposj8
==Video Solution by TheBeautyofMath==
Solved Mentally writing only the answer, and then regular way also
https://youtu.be/DE3P50S7EWw?si=HPY5cYrcZfcBpe4C
~IceMatrix


{{AMC12 box|year=2021 Fall|ab=A|num-b=21|num-a=23}}
{{AMC12 box|year=2021 Fall|ab=A|num-b=21|num-a=23}}
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 16:59, 21 September 2025

Problem

Azar and Carl play a game of tic-tac-toe. Azar places an $X$ 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 9 boxes are filled or one of the players has 3 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

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 $\binom{6}{3} - 2 = 18$. (6 open spots to put the $X$ and subtracting cases where they fill row/column)

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)

Video Solution by OmegaLearn

https://youtu.be/kxgUdv_L-ys?t=796

~ pi_is_3.14

Video Solution by Mathematical Dexterity

https://www.youtube.com/watch?v=OpRk-iposj8

Video Solution by TheBeautyofMath

Solved Mentally writing only the answer, and then regular way also

https://youtu.be/DE3P50S7EWw?si=HPY5cYrcZfcBpe4C

~IceMatrix

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
All AMC 12 Problems and Solutions

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