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

(Created page with "==Problem== Azar and Carl play a game of tic-tac-toe. Azar places an in <math>X</math> one of the boxes in a 3-by-3 array of boxes, then Carl places an <math>O</math> in one o...")
 
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==Solution==
 
==Solution==
  
'''Case 1: Carl wins with a complete row or column.'''
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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}</math> ways that Azar could have played. However, two of these ways would lead to Azar winning, so they must be excluded. This leads to <math>6\left({6 \choose 3}-2\right)=108</math> ways the board could look after the game is over.
  
Carl can win in three rows or three columns.  
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Also, there are two ways for Carl to win with a complete diagonal. For each way, there are six spaces where Azar could have played and <math>{6 \choose 3}</math> ways Azar could have played. This leads to <math>2{6 \choose 3}=40</math> ways the board could look.  
  
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.
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In total, are a total of <math>108+40 \implies \boxed{\textbf{(D) } 148}</math> ways the board could look.
 
 
This leads to <math>6({6 \choose 3}-2)=108</math> 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 <math>2{6 \choose 3}=40</math> possible board conditions.
 
 
 
Therefore, there are a total of <math>108+40 \implies \boxed{\textbf{(D) } 148}</math> board conditions.
 
  
 
{{AMC12 box|year=2021 Fall|ab=A|num-a=20|num-b=22}}
 
{{AMC12 box|year=2021 Fall|ab=A|num-a=20|num-b=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:38, 23 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

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 ${6 \choose 3}$ ways that Azar could have played. However, two of these ways would lead to Azar winning, so they must be excluded. This leads to $6\left({6 \choose 3}-2\right)=108$ 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 where Azar could have played and ${6 \choose 3}$ ways Azar could have played. This leads to $2{6 \choose 3}=40$ ways the board could look.

In total, are a total of $108+40 \implies \boxed{\textbf{(D) } 148}$ ways the board could look.

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 20
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All AMC 12 Problems and Solutions

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