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> | ||
| − | ==Solution== | + | ==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 <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. | 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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In total, there are <math>108+40 = \boxed{\textbf{(D) } 148}</math> ways the board could look after the game is over. | In total, there are <math>108+40 = \boxed{\textbf{(D) } 148}</math> ways the board could look after the game is over. | ||
| + | |||
| + | == Solution 2 == | ||
| + | We need to find out the number of configurations with 3 <math>O</math> and 3 <math>X</math> with 3 <math>O</math> in a row, and 3 <math>X</math> not in a row. | ||
| + | |||
| + | <math>\textbf{Case 1}</math>: 3 <math>O</math> are in a horizontal row or a vertical row. | ||
| + | |||
| + | Step 1: We determine the row that 3 <math>O</math> occupy. | ||
| + | |||
| + | The number of ways is 6. | ||
| + | |||
| + | Step 2: We determine the configuration of 3 <math>X</math>. | ||
| + | |||
| + | The number of ways is <math>\binom{6}{3} - 2 = 18</math>. | ||
| + | |||
| + | In this case, following from the rule of product, the number of ways is <math>6 \cdot 18 = 108</math>. | ||
| + | |||
| + | <math>\textbf{Case 2}</math>: 3 <math>O</math> are in a diagonal row. | ||
| + | |||
| + | Step 1: We determine the row that 3 <math>O</math> occupy. | ||
| + | |||
| + | The number of ways is 2. | ||
| + | |||
| + | Step 2: We determine the configuration of 3 <math>X</math>. | ||
| + | |||
| + | The number of ways is <math>\binom{6}{3} = 20</math>. | ||
| + | |||
| + | In this case, following from the rule of product, the number of ways is <math>2 \cdot 20 = 40</math>. | ||
| + | |||
| + | Putting all cases together, the total number of ways is <math>108 + 40 = 148</math>. | ||
| + | |||
| + | Therefore, the answer is <math>\boxed{\textbf{(D) }148}</math>. | ||
| + | |||
| + | ~Steven Chen (www.professorchenedu.com) | ||
{{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}} | ||
Revision as of 20:54, 25 November 2021
Problem
Azar and Carl play a game of tic-tac-toe. Azar places an in 
one of the boxes in a 3-by-3 array of boxes, then Carl places an 
in one of the remaining boxes. After that, Azar places an 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 . How many ways can the board look after the game is over?
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 
ways that Azar could have played. However, two of these ways would lead to Azar winning and must be excluded. This leads to 
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 ways Azar could have played. This leads to 
ways the board could look.
In total, there are 
ways the board could look after the game is over.
Solution 2
We need to find out the number of configurations with 3 and 3 with 3 in a row, and 3 not in a row.
: 3 are in a horizontal row or a vertical row.
Step 1: We determine the row that 3 occupy.
The number of ways is 6.
Step 2: We determine the configuration of 3 .
The number of ways is 
.
In this case, following from the rule of product, the number of ways is 
.
: 3 are in a diagonal row.
Step 1: We determine the row that 3 occupy.
The number of ways is 2.
Step 2: We determine the configuration of 3 .
The number of ways is 
.
In this case, following from the rule of product, the number of ways is 
.
Putting all cases together, the total number of ways is 
.
Therefore, the answer is 
.
~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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