Difference between revisions of "2021 Fall AMC 12A Problems/Problem 22"
Scrabbler94 (talk | contribs) (→Problem: correct the problem statement) |
|||
(3 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Azar and Carl play a game of tic-tac-toe. Azar places an | + | 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> | ||
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== | ==Video Solution by Mathematical Dexterity== | ||
https://www.youtube.com/watch?v=OpRk-iposj8 | 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 15:05, 20 February 2025
Contents
Problem
Azar and Carl play a game of tic-tac-toe. Azar places an one of the boxes in a
-by-
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 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
. How many ways can the board look after the game is over?
Solution
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)
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.