2021 AMC 10A Problems/Problem 25
Contents
- 1 Problem
- 2 Solution 1 (Casework on the Center's Color Chip's Configurations)
- 3 Solution 2 (Casework on the Top-Center and Center-Left Chips)
- 4 Solution 3 (Casework on the Red Chips' Configurations)
- 5 Solution 4 (Focusing On Top 2 Rows)
- 6 Solution 5 (Casework and Symmetry)
- 7 Solution 6 (Casework and Derangements)
- 8 Solution 7 (Rush, Use if you have less than 5 minutes)
- 9 Simple Video Solution by grogg007 (2 cases)
- 10 Video Solution (Easiest)
- 11 Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)
- 12 Video Solution by The Power of Logic
- 13 Video Solution by MRENTHUSIASM (English & Chinese)
- 14 See Also
Problem
How many ways are there to place indistinguishable red chips,
indistinguishable blue chips, and
indistinguishable green chips in the squares of a
grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
Solution 1 (Casework on the Center's Color Chip's Configurations)
Call the different colors A,B,C. There are ways to rearrange these colors to these three letters, so
must be multiplied after the letters are permuted in the grid.
WLOG assume that A is in the center.
In this configuration, there are two cases, either all the A's lie on the same diagonal:
or all the other two A's are on adjacent corners:
In the first case there are two ways to order them since there are two diagonals, and in the second case there are four ways to order them since there are four pairs of adjacent corners.
In each case there is only one way to put the three B's and the three C's as shown in the diagrams.
This means that there are
ways to arrange A,B, and C in the grid, and there are
ways to rearrange the colors. Therefore, there are
ways in total, which is
.
-happykeeper
Solution 2 (Casework on the Top-Center and Center-Left Chips)
Without the loss of generality, we fix the top-left square with a red chip. We apply casework to its two adjacent chips:
Case (1): The top-center and center-left chips have different colors.
There are three subcases for Case (1):
As there are
permutations of the three colors, each subcase has
ways. So, Case (1) has
ways in total.
Case (2): The top-center and center-left chips have the same color.
There are three subcases for Case (2):
As there are
permutations of the three colors, each subcase has
ways. So, Case (2) has
ways in total.
Answer
Together, the answer is
~MRENTHUSIASM
Solution 3 (Casework on the Red Chips' Configurations)
We consider all possible configurations of the red chips for which rotations matter:
As there are
permutations of blue and green for each configuration, the answer is
~MRENTHUSIASM (credit given to FlameKhoEmberish)
Solution 4 (Focusing On Top 2 Rows)
We will first count the colorings where if the top two rows are filled out, the grid is uniquely determined. This will only happen if there are cells of one color,
cells of another color, and
cell of the remaining color in the top
x
grid made up of cells
. With this, we are left with two tokens of one color and one token of another color for the bottom row, so there must only be
way to fill it up since the two tokens of the same color can't be next to each other. This makes the grid uniquely determined.
The cells with the same color can either be cells
or
If we choose
the two cells with the same color must be cells
and
and cell
must be the cell of the other color. If we choose
the two cells with the same color must be cells
and
and cell
must be the cell of the other color. So there are
ways to choose which cells share
colors and
way to choose which cells share
and
color. There are
ways to assign colors to the cells. We have a total of
colorings for this case. An example coloring is shown below with cells
being chosen for the
cells of the same color.
Now, what about the colorings where even if the top two rows are filled out, the grid is still not uniquely determined? In other words, what about the colorings where even if we fill in the top two rows, there is still more than one possible color combination for the last row?
In this case, the x
grid made up of the top two rows would have to use two tokens of each color, and the second row would be some permutation of the colors blue, red, green. Then for the last row, we would be left with one token of each color.
There are ways to permute the second row. WLOG, say we fill the second row of the grid so that cell
is red, cell
is blue, and cell
is green. Then cells
can be green, red, blue or blue, green, red, respectively. Finally, cells
can also either be green, red, blue or blue, green, red, respectively. This gives us
colorings.
Our final answer is .
Solution 5 (Casework and Symmetry)
There are choices for
,
choices for
.
on the down left corner can be switched with
on the upper right corner.
There are choices for
,
choices for
.
Note that is a
° rotation of
.
Note that is a
° rotation of
.
Therefore, the answer is .
Solution 6 (Casework and Derangements)
Case (1): We have a permutation of R, B, and G as all of the rows. There are ways to rearrange these three colors. After finishing the first row, we move onto the second. Notice how the second row must be a derangement of the first one. By the derangement formula,
, so there are two possible permutations of the second row. (Note: You could have also found the number of derangements of PIE). Finally, there are
possible permutations for the last row. Thus, there are
possibilities.
Case (2): All of the rows have two chips that are the same color and one that is different. There are obviously possible configurations for the first row,
for the second, and
for the third. Thus, there are
possibilities.
Therefore, our answer is
~michaelchang1
Solution 7 (Rush, Use if you have less than 5 minutes)
Ignore the center piece. Notice that when you place 3 of the chips, there are ways, making it inevitable for
ways left for the other 2, so
multiply
is
. Now, notice there are
ways to place a center piece, so our final statement is
~hashbrown2009
Simple Video Solution by grogg007 (2 cases)
https://www.youtube.com/watch?v=L5kI8cTAzyI
Video Solution (Easiest)
https://www.youtube.com/watch?v=UPUrYN1YuVA ~ MathEx
Video Solution by OmegaLearn (Symmetry, Casework, and Reflections/Rotations)
https://youtu.be/wKJ9ppI-8Ew ~ pi_is_3.14
Video Solution by The Power of Logic
https://www.youtube.com/watch?v=TEsHuvXA9Ic
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=_2hCBZHb3SA
~MRENTHUSIASM
See Also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 10 Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.