smc

This problem can be approached by Graph Theory. Note that each face of the octahedron is connected to 3 other faces. We use the above graph to represent the problem. Each vertex represents a face of the octahedron, each edge represent the octahedron's edge. Now the problem becomes how many distinguishable ways to color the $8$ vertices such that two colored graphs are distinguishable if neither can be rotated and reflected to become the other. Notice that once the outer 4 vertices are colored, no matter how the inner 4 vertices are colored, the resulting graphs are distinguishable graphs. There are $8$ colors and $4$ outer vertices, therefore there are $\left(\genfrac{}{}{0ex}{}{8}{4}\right)$ ways to color outer 4 vertices. Combination is used because the coloring has to be distinguishable when rotated and reflected. There are $4$ colors left, therefore there are $4!$ ways to color inner 4 vertices. Permutation is used because the coloring of the inner vertices have no restrictions. In total that is $\left(\genfrac{}{}{0ex}{}{8}{4}\right)\cdot 4!=\boxed{1680}$