Color 6 faces of a cube with 5 colors so that 2 faces are the same color and other colors are used once. Rotating a cube doesn't give another way. There are so many cases that I can't count so I'm stuck. It is impossible to determine the possibilities for one face and then changing the color layout since 2 faces must be the same color.
 A: Call the colours $A,B,C,D,E$ and suppose that $A$ is used twice. Note that the two $A$ faces can be positioned in only two ways. 


*

*If they are adjacent, there are $\frac{4!}2=12$ distinct ways to arrange the remaining colours – the division by two due to a 180° rotation along the edge shared by the $A$ faces.

*If they are opposite, there are only three distinct ways: there is only the choice of which colour is opposite $B$, after which all remaining colourings are equivalent.


Thus there are 15 non-equivalent ways to colour the cube with $A$ appearing twice. Multiplying by five gives 75 ways in all.
A: Just use the Burnside lemma.
The rotational group of the cube has order $24$.
Suppose a rotation other than the identity fixes a coloring, then the rotation would have $4$ fixed faces and transpose the other two, something that is clearly impossible.
Using Burnside's lemma the number of colorings is simply $\frac{5\times\binom{6}{1,1,1,1,2}}{24}=\frac{5\times6!}{2\times 24}=75$. (Because only the identity permutation fixes any colorings, and it clearly fixes all of them).
A: Here is another method of solution: apply the Polya Enumeration Theorem, also known as "Polya's Theory of Counting".  Let's say the colors are V, W, X, Y, and Z, and we want to count the cases where there are two Zs and one each of all the other colors.
As shown in the Wikipedia article on the cycle index, the cycle index of the group of face permutations of a cube is
$$\frac{1}{24} (a_1^6 + 6 a_1^2 a_4 + 3 a_1^2 a_2^2 + 8 a_3^2 + 6 a_2^3)$$
The figure inventory for our set of colors is
$$v + w + x + y + z$$
We "substitute" the figure inventory into the cycle index by replacing $a_i$ with $v^i + w^i + x^i + y^i + z^i$.  The result is
$$\frac{1}{24} [(v+w+x+y+z)^6 + 6 (v^2+w^2+x^2+y^2+z^2)^2 (v^4+w^4+x^4+y^4+z^4) + 3 (v+w+x+y+z)^2 (v^2+w^2+x^2+y^2+z^2)^2 + 8 (v^3+w^3+x^3+y^3+z^3)^2 + 6 (v^2+w^2+x^2+y^2+z^2)^3 ]$$
When this multinomial is expanded, the coefficient of $v w x y z^2$ is $15$.  (I used a computer algebra system, but I don't think it would be very hard to do a pencil and paper solution.)  
So the number of ways to paint the faces of a cube with two Zs and one each of V, W, X, and Y is 15.
