Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 Fall AMC 12A Problems/Problem 22"

m
m (→‎Video Solution by TheBeautyofMath)
 
(8 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
==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.
  
−
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.
+
<math>\textbf{Case 1}</math>: 3 <math>O</math> are in a horizontal row or a vertical row.
  
−
Also, there are two ways for Carl to win with a complete diagonal. For each way, there are six spaces 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.  
+
Step 1: We determine the row that 3 <math>O</math> occupy.
  
−
In total, there are <math>108+40 \implies \boxed{\textbf{(D) } 148}</math> ways the board could look.
+
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)
 +
 
 +
== 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}}
 
{{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

Problem

Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a $3$-by-$3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ 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 $O$. How many ways can the board look after the game is over?

$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$

Solution

We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.

$\textbf{Case 1}$: 3 $O$ are in a horizontal row or a vertical row.

Step 1: We determine the row that 3 $O$ occupy.

The number of ways is 6.

Step 2: We determine the configuration of 3 $X$.

The number of ways is $\binom{6}{3} - 2 = 18$.

In this case, following from the rule of product, the number of ways is $6 \cdot 18 = 108$.

$\textbf{Case 2}$: 3 $O$ are in a diagonal row.

Step 1: We determine the row that 3 $O$ occupy.

The number of ways is 2.

Step 2: We determine the configuration of 3 $X$.

The number of ways is $\binom{6}{3} = 20$.

In this case, following from the rule of product, the number of ways is $2 \cdot 20 = 40$.

Putting all cases together, the total number of ways is $108 + 40 = 148$.

Therefore, the answer is $\boxed{\textbf{(D) }148}$.

~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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12A_Problems/Problem_22&oldid=229610"