Mônica has six colors to paint the faces of a cube. In how many ways can she do this if:

a) All faces have the same color?

b) Five faces have the same color and one face has a different color from the rest?

c) All faces have different colors?

Note: remember, for example, that the two paintings below are the same, since if we rotate one of them appropriately, we will get the other!
![](https://cdn.mathpix.com/cropped/2024_05_01_d18780f0fd52e0502cf8g-03.jpg?height=384&width=808&top_left_y=798&top_left_x=724)
Solution

a) Since all faces must be the same color, we can only paint the die with one color. With six colors to choose from, there are six distinct ways to paint the cube.

b) Paint one face with one of the six possible colors. There are six possibilities for this. For each of these possibilities, the remaining faces can be painted, all with the same color, from one of the five remaining colors. Therefore, we have $6 \times 5=30$ possibilities. Note that it doesn't matter which face was initially chosen, as we can always rotate the cube in an appropriate manner to show that the paintings are the same.
c) If we paint the cube in a certain position, we will have: six possibilities for the first face; five for the second; four for the third; three for the fourth; two for the fifth; and one for the last, which gives $6 \times 5 \times 4 \times 3 \times 2 \times 1=720$. However, in this method, we are counting the same painting multiple times, as rotating the cube does not change the painting. Let's count how many times we are counting each painting repeatedly. Fix one color, say blue. This blue face can be placed in six different positions. In each of these positions, we can rotate the cube into four distinct positions (keeping the blue face fixed). Therefore, we conclude that each painting is counted $6 \times 4=24$ times. Thus, 720 divided by 24 gives 30 different ways to paint the cube.