+36 Each vertex of cube a b c d e f g h is colored either black or white. Find the number of colorings in which all four vertices of some face have the same color.
+411 To count the number of colorings where all four vertices of some face have the same color, we can consider cases based on the number of faces with all vertices the same color:
1. One face with all vertices the same color:
Choose one of the six faces to have all vertices the same color (2 choices: black or white).
The remaining five faces can be colored arbitrarily (2^5 possibilities).
Total: 2 * 2^5 = 64
2. Two faces with all vertices the same color:
Choose two faces with the same color (2 choices: both black or both white).
The remaining four faces can be colored arbitrarily (2^4 possibilities).
However, we need to avoid double-counting cases where the two faces share an edge. There are 12 edges in a cube, and each edge is shared by two faces. So, there are 12/2 = 6 cases to avoid.
Total: 2 * 2^4 - 6 = 26
3. Three faces with all vertices the same color:
Choose three faces with the same color (2 choices: all black or all white).
The remaining three faces can be colored arbitrarily (2^3 possibilities).
However, we need to avoid double-counting cases where two or three faces share an edge. There are 12 edges in a cube, and each edge is shared by two faces. So, there are 12/2 = 6 cases to avoid.
Total: 2 * 2^3 - 6 = 2
4. Four faces with all vertices the same color:
There are four colorings in this case.
Therefore, the total number of colorings is 64 + 26 + 2 + 4 = 96.
