Can I have a hint on this combinatorics question? Six faces of identical cubes are painted 6 distinct colors. How many different cubes can be formed?
The answer is surprisingly small but that may be because I don't know how to express the different ways properly. I just incorrectly did $6!$ because if there are 6 ways to color the first face, there are now 5 colors left so 5 ways to color the second face. And so on. Why is this wrong?
Perhaps I need to make use of "identical" and "distinct" from the question. If the cubes are identical, I must divide the answer by something, perhaps $6!$ for the 6 identical faces? And there should be some "choosing" happening but I can see where this fits in but it must be for the "distinct" part. Choose colors as in $6C1, 5C1 ...etc$ but this is the same as $6!$
Please help
 A: Two cubes that differ by a rotation are considered the same. There are $24$ ways to orient the cube, all of which are distinct because of the use of distinct colours, so $6!$ must be divided by $24$ to get the correct answer of $30$.
A: This is perhaps a more concrete way to arrive at the answer.  Say we will color the faces with the colors, $R,O,Y,G,B,V$.  We have to color some face $R$.  Place the cube on the table, with that face down.  Now we have to color some face $O$.
We can paint the top face, or one of the lateral faces.  Suppose first that we paint the top face.  Now we have to paint one of the lateral faces $Y$.  Whichever face we choose, we can rotate the cube so that that face is facing us.  Now the cube is fixed in place.  $R$ on bottom, $O$ on top, $Y$ facing us.  We can't rotate it without changing the arrangement.   There are $3!$ ways to paint the remaining $3$ faces.
If instead, we paint one of the lateral faces $O$, we can rotate the cube so $O$ is facing us, and again we find it is fixed in place; $R$ on bottom, $O$ facing us.  We have $4!$ ways to paint the remaining $4$ faces.
You may be interpreting the question a bit too literally.  There are indeed $6!$ orders in which we can choose to color the faces of the cube, but in many cases, the results are indistinguishable.  The question is asking for the number of distinct cubes that can be made by coloring each face with one of $6$ given colors.
A: 
If you label the cube counterclockwise $1$ through $4$ starting from the yellow face, the top face $5$, the bottom $6$, then if we are given the colors: yellow, blue, orange, clear, red and green, pretending the front most face is clear/see-through, a coloring of the cube would be a permutation: $\{1,2,3,4,5,6\}\rightarrow$ {yellow, blue, orange, clear, red, green}.
There are $6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\cdot = 6!$ ways to color this cube. But now every coloring can be described the same way $6\cdot 4 =24$ 'different ways'. Looking at our cube, if we rotate it front, back, left, right, enough times we can have $6$ distinct bottom faces, the top face is automatically determined once the bottom face is. Next given a bottom face, we can rotate the cube $90^{\circ}$ counterclockwise 4 times and still have the same coloring.
A: Perhaps to record a more advanced answer for those who knows Burnside lemma from group theory:
Denote $X$ to be the set of all possible ways to label each of the 6 faces of the cube with exactly one of six distinct colors. There are $6!$ ways of doing this without considering which ones are the same by rotation.
Now the rotational symmetry group for a cube is the octahedral group $O \simeq S_4$ (https://en.wikipedia.org/wiki/Octahedral_symmetry) with $|O| = 24$.
Now the group $O$ acts on the set $X$, and in each orbit it contains all the colorings that are considered the same when up to rotation. By Burnside lemma, we have the number of orbits $|X/O| = \frac{1}{|O|}\sum_{g\in O} |X^g|$, where $X^g$ is the set of colorings that are unchanged by the action of $g$. Since the cube has six different colors, whenever $g\neq e$ we have $|X^g|=0$, as any nontrivial rotation will not fix the coloring configuration. Unless $g=e$, then we have $|X^e|=|X|= 6!$, as every configuration is fixed by identity.
Hence $|X/O| = \frac{1}{|O|}\sum_{g\in O} |X^g| = \frac{6!}{24}=30$.
