Possible ways to distinctly paint a cube

Question

We had a mathematics competition recently, consisting of 5 questions. The last question was especially tricky.

The exact wording of the question was:

"Philippe is painting cubes. Every cube he paints must have each of its six faces painted exactly one color, either red, yellow, or blue, and he must use all three color on each cube. Two cubes have distinct paint jobs if there is no way to place the two painted cubes on a table so that all pairs of corresponding faces (ie., fronts backs, tops, bottoms, lefts, rights) are the same color. Find the maximum number of cubes that Philippe can paint given that each cube’s paint job is distinct from all other cubes’ paint jobs:"

Of the hundreds of students present at the competition, only one was able to get it correct.

A couple of friends and I have tried solving the question using various different methods, but we have not succeeded with a single iteration.

My attempt went something like this:

• We can represent each side of the cube as an element of the set $ \{a, b, c, d, e, f\}$, and we can represent the possible paint colors with R for red, Y for yellow, or B for blue.

• Out of these 6 sides, 1 must be R, 1 must be Y, and 1 must be B

• Thus, out of the $a, b, c, d, e, f$, there must exist some string that is a combination of $x, y, z$ that has the states $R, Y,$ and $B$

• This string (of length $3$) must be pulled from the original string $a, b, c, d, e, f$ (of length $6$), and thus, can be found as $6 \choose 3$ which evaluates to $20$ possible combinations of $x, y, z$ choosen from $a, b, c, d, e, f$

• For each $x, y, z$, there are still $3$ more values of $a, b, c, d, e, f$ that are undecided, and for each of these $3$ values, there are $3$ possible states (R, Y, B), so thus, there 27 different variations of each $x, y, z$.

• Since there are $20$ such strings $x, y, z$, and there are 27 variations of each $x, y, z$, the total amount of variations becomes 20 × 27 = 540. Let each of these variations have the form "$q, w, e, r, t, y$"

• In this problem, we are trying to find distinct solutions, which are ineffected by position, we can find the other positions of "$q, w, e, r, t, y$" to be translations of itself. For example, "$w, e, r, t, y, q$" is congruent to "$q, w, e, r, t, y$". For each "$q, w, e, r, t, y$", there exist $5$ alternate translated states, for a total of 6 possibilities for each "$q, w, e, r, t, y$".

• Thus, since there is only 1 true variation for every $6$ out of the $540$ total variations, the true total number of variations should $\dfrac{540}{6}$ which is 90.

I (and all my friends who reached the same solution), were incorrect. The true solution was $30$. We have tried to go over our work many times but we can't find the error in our work.

What we did come to a conclusion on was that $30$ is equal to $\dfrac{540}{18}$, and we had gotten $\dfrac{540}{6}$. Thus, we assumed that somewhere in our work, we has missed a $3$, and thus, instead of $\dfrac{540}{6×3}$, we simply got $\dfrac{540}{6}$.

If someone can point out our error, it would greatly help.

Tips on how to approach questions such as this in the future would also be greatly appreciated.