Difference between revisions of "2021 Fall AMC 12A Problems/Problem 22"
Ehuang0531 (talk | contribs) (→Solution) |
Ehuang0531 (talk | contribs) m |
||
| Line 12: | Line 12: | ||
In total, there are <math>108+40 \implies \boxed{\textbf{(D) } 148}</math> ways the board could look. | In total, there are <math>108+40 \implies \boxed{\textbf{(D) } 148}</math> ways the board could look. | ||
| − | {{AMC12 box|year=2021 Fall|ab=A|num-a= | + | {{AMC12 box|year=2021 Fall|ab=A|num-a=21|num-b=23}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:42, 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 and ways Azar could have played. This leads to 
ways the board could look.
In total, there are 
ways the board could look.
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 21 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
