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)

We only need to consider the arrangement of the cubes of one color (say, white), as the four other cubes will be all of the other color (say, black).

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

Case 1: Each layer contains 2 white cubes.

There are 2 ways the white cubes in each layer can be arranged: the white cubes can be adjacent or diagonal to each other and there are $2 \cdot 2 = 4$ ways two of such layers can be arranged.

Case 2: One layer contains 1 white cube and the other layer contains 3 white cubes.

Assume that we view the cube by rotating it such that there is a white cube in the upper-left of the front layer. If the sole black cube on the back layer is also on the upper-left, then if we split this into left and right layers, the right layer has only two white cubes and is covered in case 1. The other three places for the black cube make it such that no matter how we separate the cube into two layers, each layer will always contain one cube of one color and three cubes of another color. So, we add on another 3 ways the $2 \times 2 \times 2$ cube could be.

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.

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

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

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2021 Fall AMC 12B Problems/Problem 20

Problem

Solution 1 (Direct Counting)

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