2022 AMC 8 Problems/Problem 23

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. $[asy] //diagram size(5cm); defaultpen(linewidth(1.5)); real r = 0.37; path equi = r * dir(-30) -- (r+0.03) * dir(90) -- r * dir(210) -- cycle; draw((0,0)--(0,3)--(3,3)--(3,0)--cycle); draw((0,1)--(3,1)--(3,2)--(0,2)--cycle); draw((1,0)--(1,3)--(2,3)--(2,0)--cycle); draw(circle((3/2,5/2),1/3)); draw(circle((5/2,1/2),1/3)); draw(circle((3/2,3/2),1/3)); draw(shift(0.5,0.38) * equi); draw(shift(1.5,0.38) * equi); draw(shift(0.5,1.38) * equi); draw(shift(2.5,1.38) * equi); draw(shift(0.5,2.38) * equi); draw(shift(2.5,2.38) * equi); [/asy]$ 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

We are given a 3x3 grid in the problem. Therefore, there are 9 possible spots for a triangle or circle to be placed, and we are then asked to choose any 3 possible positions in a row for either a triangle or circle. Thus we have $\binom{9}{3}$ which is equivalent to $\boxed{\textbf{(D) }84}$

~Clew28

Solution 2 (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. (We subtract $2$ from the last line because one arrangement of the last line is all triangles and the other arrangement of the last line is all circles, which causes Case 2 to overlap with Case 1 and further complicating the solution.) In total, this is $6\cdot 6 = 36.$

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

~wamofan

Solution 3

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 by Math-X (First understand the problem!!!)

I made A SECOND VERSION ( very easy to understand) https://youtu.be/ukCWuMbxxLU

~Math-X

~ pi_is_3.14

Video Solution

~Mathematical Dexterity

~Interstigation

~David

~STEMbreezy

~savannahsolver