, 2014

Solution:

Answer: $35$

We do casework on $R$, the number of red vertices. Let the cube be called $A B C D E F G H$, with opposite faces $A B C D$ and $E F G H$, such that $A$ is directly above $E$.

- $\underline{R=0}$ : There is one such coloring, which has only blue vertices.

- $\underline{R=1}$ : There are $8$ ways to choose the red vertex, and all other vertices must be blue. There are $8$ colorings in this case.

- $\underline{R=2}$ : Any pair not an edge works, so the answer is $\binom{8}{2}-12=16$.

- $\underline{R}=3$ : Each face $A B C D$ and $E F G H$ has at most two red spots. Assume WLOG $A B C D$ has exactly two and $E F G H$ has exactly one (multiply by $2$ at the end). There are two ways to pick those in $A B C D$ (two opposite corners), and two ways after that to pick $E F G H$. Hence the grand total for this subcase is $2 \cdot 2 \cdot 2=8$.

- $\underline{R=4}$ : There are only two ways to do this.

Hence, the sum is $35$.