Difference between revisions of "2022 AMC 8 Problems/Problem 23"
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==Solution 1 (Casework)== | ==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. | |
− | . In total, this is <math>6\cdot 6 = 36.</math> | + | We take casework: |
+ | |||
+ | '''Case 1: 3 lines''': | ||
+ | In this case, the lines would need to be <math>2</math> of one shape and <math>1</math> of another, so there are <math>\frac{3!}{2} = 3</math> ways to arrange the lines and <math>2</math> ways to pick which shape has only one line. In total, this is <math>3\cdot 2 = 6.</math> | ||
+ | |||
+ | '''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 <math>3! = 6</math> ways to arrange the lines and <math>2^3-2 = 6</math> ways to choose the last line. In total, this is <math>6\cdot 6 = 36.</math> | ||
Finally, we add and multiply: <math>2(36+6)=2(42)=\boxed{\textbf{(D) }84}</math>. | Finally, we add and multiply: <math>2(36+6)=2(42)=\boxed{\textbf{(D) }84}</math>. |
Revision as of 12:30, 15 January 2023
Contents
Problem
A or
is placed in each of the nine squares in a
-by-
grid. Shown below is a sample configuration with three
s in a line.
How many configurations will have three
s in a line and three
s in a line?
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 of one shape and
of another, so there are
ways to arrange the lines and
ways to pick which shape has only one line. In total, this is
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 ways to arrange the lines and
ways to choose the last line. In total, this is
Finally, we add and multiply: .
~wamofan
Solution 2
We will only consider columns, but at the end our answer should be multiplied by . There are
ways to choose a column for
and
ways to choose a column for
. The third column can be filled in
ways. Therefore, we have
ways. However, we overcounted the cases with
complete columns of with one symbol and
complete column with another symbol. This happens in
ways.
. However, we have to remember to double our answer giving us
.
~MathFun1000
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
Video Solution
https://youtu.be/0orAAUaLIO0?t=257
~STEMbreezy
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.