HMMT-Feb 2024 comb, Problem 3

Compute the number of ways there are to assemble 2 red unit cubes and 25 white unit cubes into a $3 \times 3 \times 3$ cube such that red is visible on exactly 4 faces of the larger cube. (Rotations and reflections are considered distinct.)

Solution: \section*{Solution:} We do casework on the two red unit cubes; they can either be in a corner, an edge, or the center of the face. \begin{itemize}   \item If they are both in a corner, they must be adjacent - for each configuration, this corresponds to an edge, of which there are 12 . \end{itemize} \begin{itemize}   \item If one is in the corner and the other is at an edge, we have 8 choices to place the corner. For the edge, the red edge square has to go on the boundary of the faces touching the red corner square, and there are six places here. Thus, we get $8 \cdot 6=48$ configurations. \end{itemize} \begin{itemize}   \item If one is a corner and the other is in the center of a face, we again have 8 choices for the corner and 3 choices for the center face (the faces not touching the red corner). This gives $8 \cdot 3=8+8+8=24$ options. \end{itemize} \begin{itemize}   \item We have now completed the cases with a red corner square! Now suppose we have two edges: If we chose in order, we have 12 choices for the first cube. For the second cube, we must place the edge so it covers two new faces, and thus we have five choices. Since we could have picked these edges in either order, we divide by two to avoid overcounting, and we have $12 \cdot 5 / 2=30$ in this case. \end{itemize} Now, since edges and faces only cover at most 2 and 1 face respectively, no other configuration works. Thus we have all the cases, and we add: $12+48+24+30=114$. $\fbox{114}$.

Answer: 114