Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?
$\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ 18 \qquad \mathrm{(C)}\ 27 \qquad \mathrm{(D)}\ 54 \qquad \mathrm{(E)}\ 81$
A tetrahedron has 4 sides.
The ratio of the number of faces with each color must be one of the following:
$4:0:0$, $3:1:0$, $2:2:0$, or $2:1:1$

The first ratio yields $3$ appearances, one of each color.
The second ratio yields $3\cdot 2 = 6$ appearances, three choices for the first color, and two choices for the second.
The third ratio yields $\binom{3}{2} = 3$ appearances since the two colors are interchangeable.
The fourth ratio yields $3$ appearances. There are three choices for the first color, and since the second two colors are interchangeable, there is only one distinguishable pair that fits them.
The total is $3 + 6 + 3 + 3 = 15$ appearances $\Rightarrow \mathrm{(A)}$