2021 AMC Fall 10B Problem 24

The big cube has $8$ corner spots, and we choose $4$ of them to be white. Turning the cube does not make a new construction. To avoid counting the same thing twice, always turn the cube so that the face with the most white cubes is on top, and sort by that number.

Case 1: a face with all 4 white. Put it on top. The top layer is white and the bottom layer is blue. That is $1$ way.

Case 2: the most on any face is 3. Put that face on top. Call the one blue top cube $D$, the top cube diagonally across from it $B$, and the other two top cubes $A$ and $C$. So the white top cubes $A, B, C$ form an L. The fourth white cube is somewhere in the bottom layer, under one of the four top cubes:

1. Under $D$. It touches $D$ and the bottom cubes under $A$ and $C$, which are all blue. So it sits alone: an L plus a lonely cube. $1$ way.
2. Under $B$. Now $B$ touches three white cubes ($A$, $C$, and the one below). It looks like a corner with three arms. $1$ way.
3. Under $A$ or under $C$. The four white cubes form a chain $C \to B \to A \to$ (below $A$), or $A \to B \to C \to$ (below $C$). Each chain turns in all three directions, like a twisted staircase. The two chains are mirror images, one twisting left and one twisting right. Turning a real object never changes which way it twists (think of a left and right glove), so these are $2$ different ways.

Case 2 total: $4$ ways.

Case 3: every face has exactly 2 white. (A face with $0$ or $1$ white would leave $4$ or $3$ white on the opposite face, which is a case already done.) Look at the two white cubes on top.

1. They are side by side. The side face that contains both of them already has $2$ white, so the two cubes under them are blue. The bottom white cubes are the other two, side by side. The shape is two white "bars" on opposite edges of the cube. $1$ way.
2. They are diagonal. Each side face already has $1$ white from the top, so the bottom whites are also diagonal. One choice puts them right under the top ones, which is the "two bars" shape again (turned on its side). The other choice puts them under the blue top cubes, so no two white cubes touch at all. That is $1$ new way.

Case 3 total: $2$ ways.

Total: $1 + 4 + 2 = 7$.

The answer is $\boxed{\textbf{(A)}\ 7}$.

When counting things that can be turned, pick a standard way to hold each one (here: the face with the most white cubes on top) and then list by shape, since turning never changes which cubes touch. The one catch in 3D is mirror images: a shape and its mirror copy count as different unless some turn matches them. As a check, the seven shapes can be placed in $6 + 24 + 8 + 12 + 12 + 6 + 2 = 70$ ways, which equals the number of ways to choose $4$ of $8$ spots.