Difference between revisions of "2021 Fall AMC 12A Problems/Problem 22"
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==Problem== | ==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 of the remaining boxes. After that, Azar places an <math>X</math> 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 <math>O</math>. How many ways can the board look after the game is over? | + | Azar and Carl play a game of tic-tac-toe. Azar places an in <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 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 <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> | ||
− | ==Solution== | + | == Solution == |
+ | 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. | |
− | {{AMC12 box|year=2021 Fall|ab=A|num- | + | 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) | ||
+ | |||
+ | == 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 | ||
+ | |||
+ | {{AMC12 box|year=2021 Fall|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 21:49, 11 October 2024
Contents
Problem
Azar and Carl play a game of tic-tac-toe. Azar places an in 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 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
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 |
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