Vertices of Cube coloured with black or white colour In how many rotionally distinct ways can the vertices of a cube be coloured with black or white colour ?
I don't know how to approach this question. Please provide some insight.
 A: This answers the wrong question! The question was about vertices, and I thought it was faces. This goes to show, people: Read through your problems thuroughly before answering them, both on this site and on assignments and tests.
The standard solution (at least if you allow a bit more complicated notations and concepts) is to use Burnside's lemma (there is a simpler, slightly less systematic solution toward the bottom of this answer if you feel intimidated by the wiki article).
The lemma states how many of the $2^6$ total possible colorings of the cube that are considered rotationally distinct. The formula, $|X/G| = \frac1{|G|}\sum_{g \in G}|X^g|$ might look a bit mysterious if you're not used to the notation, but I will try to explain it in more elementary means than the wiki article.
First of all, $|X/G|$ is the number you're after. $X$ is the set containing the $2^6$ possible colorings of the cube, $G$ is the group (or set if you will) containing the $24$ possible rotations of a cube, and $X/G$ is the set of equivalence classes of $X$, where one coloring is seen as equivalent to another if there is a rotation that takes you from one to the other.
The other side of the formula contains the fraction $\frac{1}{|G|}$, where $|G| = 24$ is the number of possible rotations of a cube (including the trivial "doing nothing" rotation). Then for each possible rotation $g \in G$ we find the set $X^g$, which is the set of colorings which look exactly the same after applying $g$. The last step is simply to add the different sizes $|X^g|$ and divide by $24$.
So we do this systematically by $g$, seeing how many different colorings are completely unchanged by applying $g$. There are a few different "types" of $g$'s to consider. See https://www.youtube.com/watch?v=gBg4-lJ19Gg for an illustration of the different rotations.


*

*The trivial rotation. There is only one of those, and in that case $|X^g| = 2^6$

*A $90^\circ$ face rotation. There are six of these (axes $1, 2, 3$ in the video, either $90^\circ$ clockwise or antoclockwise). In this case, the face turned, and the one opposite it can be any color they want, but for $g$ to not change what the remaining four faces look like, they all need to have the same color. Therefore $|X^g| = 2^3$ in this case.

*A $180^\circ$ face rotation. There are three of these (same axes, but $180^\circ$, so clockwise or anticlockwise doesn't matter). Once again, the face you rotate and the one opposite it can be whatever you like, but now the remaining four are paired up into two independent pairs. Therefore, in this case, $|X^g| = 2^4$.

*A $120^\circ$ corner twist. There are $8$ of these (axes $4, 5, 6$ and $7$, each either one or two steps). They each group together two sets of three faces that much have the same color. Thus $|X^g| = 2^2$.

*Lastly, the $180^\circ$ rotations described by axes $8$ through $13$ in the video (I don't know a simple way to describe using words). That makes $6$ different $g$'s, and each one pairs up three sets of two sides that must be the same, so $|X^g| = 2^3$.


Finally, we may add all the different $|X^g|$'s together to get
$$
\sum_{g \in G}|X^g| = 2^6 + 6\cdot 2^3 + 3\cdot 2^4 + 8\cdot 2^2 + 6\cdot 2^3 = 1240
$$
Finally, dividing my $24$ to complete the formula gives an answer of $10$.

Of course, this specific problem is so simple that Burnside's lemma is probably overkill. I'll do a direct count too, just to show how to do that. I will order it by the number of white faces.


*

*There is one way to color the cube using no white faces (rotationally invariant or not)

*There is one rotationally invariant way of painting the cube with one white face.

*There are two ways of painting the cube using two white faces. One where the white faces are adjacent, and one where they are opposite.

*There are two ways to paint a cube using three white faces. Either you have the three faces clustered together adjacently, or you've covered a "band", with two of the white faces opposite and one white face between them.

*There are two ways of painting a cube with four white faces. Either the remaining two black faces are adjacent, or they are opposite.

*There is one way to paint the cube using five white faces (and one black face).

*There is one way to paint the cube all white.


And we count $10$ ways in total.
A: You can still use Burnside's Lemma for coloring the vertices, because the group of symmetric rotations of a cube (where |G| = 24) acts on the set of vertices, as well as the set of faces.
The "identity rotation" leaves all 2^8 possibilities untransformed so that will be the first term in the summation,  and then the total will be divided by 24, the number of symmetric rotations on a cube.
So (16^2 + 6 * 4 + 3 * 16 + 8 * 16 + 6 * 16)/24
= 1/3 * (32+3+6+16 +12) = 69/3 = 23
(unless I've  counted wrong or done the math wrong,  but since I came up with an integer, I tend to think that I got it right. )
