Difference between revisions of "2021 Fall AMC 12A Problems/Problem 22"
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==Solution== | ==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 <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 | + | 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, so they must be excluded. This leads to <math>6(20-2)=108</math> 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 <math>{6 \choose 3}</math> ways Azar could have played. This leads to <math>2 | + | 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}=20</math> ways Azar could have played. This leads to <math>2\cdot20=40</math> ways the board could look. |
In total, are a total of <math>108+40 \implies \boxed{\textbf{(D) } 148}</math> ways the board could look. | In total, are a total of <math>108+40 \implies \boxed{\textbf{(D) } 148}</math> ways the board could look. | ||
Revision as of 19:39, 23 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
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, so they 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 where Azar could have played and ways Azar could have played. This leads to 
ways the board could look.
In total, are a total of 
ways the board could look.
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 20 |
| 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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