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Difference between revisions of "2022 AMC 8 Problems/Problem 23"

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==Solution 2==
 
==Solution 2==
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We will only consider columns, but at the end our answer should be multiplied by <math>2</math>. There are <math>3</math> ways to choose a column for <math>\bigcirc</math> and <math>2</math> ways to choose a column for <math>\triangle</math>. The third column can be filled in <math>2^3=8</math> ways. Therefore, we have <math>3\cdot2\cdot8=48</math> ways. However, we overcounted the cases with <math>2</math> complete columns of with one symbol and <math>1</math> complete columns with another symbol. This happens in <math>2\cdot3=6</math> ways. <math>48-6=42</math>. However, we have to remember to double our answer giving us <math>\boxed{\textbf{(D) }84}</math>.
+
We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are <math>3</math> ways to choose a column with all <math>\bigcirc</math>'s and <math>2</math> ways to choose a column with all <math>\triangle</math>'s. The third column can be filled in <math>2^3=8</math> ways. Therefore, we have a total of <math>3\cdot2\cdot8=48</math> cases. However, we overcounted the cases with <math>2</math> complete columns of with one symbol and <math>1</math> complete column with another symbol. This happens in <math>2\cdot3=6</math> cases. <math>48-6=42</math>. However, we have to remember to double our answer, giving us <math>\boxed{\textbf{(D) }84}</math> ways to complete the grid.
  
 
~MathFun1000
 
~MathFun1000
 +
==Video Solution==
 +
https://youtu.be/p7UHadjWqLg
 +
 +
Please like and subscribe!
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/fL7DKXZjmAo?t=239
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~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
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~Mathematical Dexterity
 
~Mathematical Dexterity
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==Video Solution==
 
==Video Solution==
 
https://youtu.be/Ij9pAy6tQSg?t=2250
 
https://youtu.be/Ij9pAy6tQSg?t=2250
  
 
~Interstigation
 
~Interstigation
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=KYglbGTvfsY 
 +
 +
~David
 +
 +
==Video Solution==
 +
https://youtu.be/0orAAUaLIO0?t=257
 +
 +
~STEMbreezy
 +
 +
==Video Solution==
 +
https://youtu.be/YYvbTopjB1E
 +
 +
~savannahsolver
 +
 +
== Video Solution==
 +
https://youtu.be/p7UHadjWqLg
 +
 +
Please like and subscribe!
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=22|num-a=24}}
 
{{AMC8 box|year=2022|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:53, 15 April 2023

Problem

A $\triangle$ or $\bigcirc$ is placed in each of the nine squares in a $3$-by-$3$ grid. Shown below is a sample configuration with three $\triangle$s in a line. [asy] //diagram by kante314 size(3.3cm); defaultpen(linewidth(1)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy] How many configurations will have three $\triangle$s in a line and three $\bigcirc$s in a line?

$\textbf{(A) } 39 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 84 \qquad \textbf{(E) } 96$

Solution 1 (Casework)

Notice that diagonals and a vertical-horizontal pair can never work, so the only possibilities are if all lines are vertical or if all lines are horizontal. These are essentially the same, so we'll count up how many work with all lines of shapes vertical, and then multiply by 2 at the end.

We take casework:

Case 1: 3 lines: In this case, the lines would need to be $2$ of one shape and $1$ of another, so there are $\frac{3!}{2} = 3$ ways to arrange the lines and $2$ ways to pick which shape has only one line. In total, this is $3\cdot 2 = 6.$

Case 2: 2 lines: In this case, the lines would be one line of triangles, one line of circles, and the last one can be anything that includes both shapes. There are $3! = 6$ ways to arrange the lines and $2^3-2 = 6$ ways to choose the last line. In total, this is $6\cdot 6 = 36.$

Finally, we add and multiply: $2(36+6)=2(42)=\boxed{\textbf{(D) }84}$.

~wamofan

Solution 2

We will only consider cases where the three identical symbols are the same column, but at the end we shall double our answer as the same holds true for rows. There are $3$ ways to choose a column with all $\bigcirc$'s and $2$ ways to choose a column with all $\triangle$'s. The third column can be filled in $2^3=8$ ways. Therefore, we have a total of $3\cdot2\cdot8=48$ cases. However, we overcounted the cases with $2$ complete columns of with one symbol and $1$ complete column with another symbol. This happens in $2\cdot3=6$ cases. $48-6=42$. However, we have to remember to double our answer, giving us $\boxed{\textbf{(D) }84}$ ways to complete the grid.

~MathFun1000

Video Solution

https://youtu.be/p7UHadjWqLg

Please like and subscribe!

Video Solution by OmegaLearn

https://youtu.be/fL7DKXZjmAo?t=239

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=or4pKVzQ3gI

~Mathematical Dexterity

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=2250

~Interstigation

Video Solution

https://www.youtube.com/watch?v=KYglbGTvfsY

~David

Video Solution

https://youtu.be/0orAAUaLIO0?t=257

~STEMbreezy

Video Solution

https://youtu.be/YYvbTopjB1E

~savannahsolver

Video Solution

https://youtu.be/p7UHadjWqLg

Please like and subscribe!

See Also

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 AJHSME/AMC 8 Problems and Solutions

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