2021 Fall AMC 12B Problems/Problem 20

The following problem is from both the 2021 Fall AMC 12B #20 and 2021 Fall AMC 12B #24, so both problems redirect to this page.

Problem

A cube is constructed from $4$ white unit cubes and $4$ black unit cubes. How many different ways are there to construct the $2 \times 2 \times 2$ cube using these smaller cubes? (Two constructions are considered the same if one can be rotated to match the other.)

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 11$

Solution 1 (Direct Counting)

Divide the $2 \times 2 \times 2$ cube into two layers.

Case 1: Each layer contains 2 cubes of each color. Note that we only need to consider the layout of the white cubes because all the other cubes will be black cubes. There are 2 ways that the two white cubes can be arranged within each layer: adjacent or diagonal to each other.
1. Case 1.1: Both layers have two white cubes adjacent to each other. Rotate the cube such that there are white cubes along the top of the front layer. Now, the white cubes in the back layer can be on the top, bottom, right, or left.^{See note 1.} There are $4$ constructions resulting from this case.
2. Case 1.2: One layer has two white cubes adjacent to each other, while the other has two white cubes diagonal from each other. Rotate the cube such that there are white cubes along the top of the front layer. The white cubes in the back layer can be at the top-left and bottom-right or the top-right and bottom-left. If we rotate the latter case by 90 degrees clockwise, we see that it matches the former case. There is $1$ additional construction resulting from this case.
3. Case 1.3: Both layers have white cubes diagonal from each other. Rotate the cube such that there is a white cube at the top-left of one layer. Therefore, there must be a white cube at the bottom-right of that layer. The other layer could also have white cubes at the top-left and the bottom-right. This is the same as case 1.1 with the white cubes in the back layer at the bottom. The other layer could have white cubes at the top-right and the bottom-left. This is a distinct case. So there is $1$ additional construction resulting from this case.
4. So case 1 results in $4+1+1=6$ arrangements.
Case 2: Each layer contains 3 cubes of one color and 1 cube of the other color. Split this cube into two layers; the sole white cube on one layer must be on the opposite corner of the sole black cube on the other layer, otherwise there will be some way to spit the cube into two layers such that there are 2 cubes of each color on each layer. So case 2 results in $1$ additional arrangement.
Case 3: One layer contains 0 white cubes and the other layer contains 4 white cubes. Only 1 possible $2 \times 2 \times 2$ cube can result from this case. However, if we divide up this $2 \times 2 \times 2$ cube in the direction perpendicular to the way we first did, then we see this is a repeat of the case where we have 2 white cubes in the first layer and the second layer is arranged such that whites and blacks are straight with each other. So there are no additional constructions resulting from this case.

Therefore, our answer is $6+1=\boxed{\textbf{(A)}\ 7}$ .

Notes

1: To prove the 3rd and 4th cases distinct, we can model them with our hands. Extend our thumbs and pointer fingers into an L. These fingers represent the three white cubes on the top layer. Our left and right hands represent the 3rd and 4th cases respectively. The 4th white cube in each case extends down from the tip of each pointer finger towards the rest of each hand. If we overlap our thumbs and pointer fingers, then the 4th cube in each situation will extend outwards in opposite directions, so these cases are distinct.

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AMC 12 Problems and Solutions

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AMC 10 Problems and Solutions

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2021 Fall AMC 12B Problems/Problem 20

Contents

Problem

Solution 1 (Direct Counting)

Notes

See Also