Let's first consider the number of ways to choose the side that is going to be painted blue. Yes. Since the cube has six sides, we have six options for the side that we are going to paint blue. Now, let's consider the number of ways to choose the two sides that are going to be painted red. After we paint one side blue, there are five sides left. So we have $5\choose{2}$ = 10 ways to choose the two sides that we are going to paint red. Now the remaining three sides must be painted green. That's correct. So we don't need to consider how to paint those sides. So the total number of ways to paint the cube is just the product of the number of ways to choose each color. That's correct. So the total number of ways is $6 * \binom{5}{2} = 60$. That's right. In the beginning, we said that two cubes are considered the same if one can be rotated to match the other. Right. So we have to divide the total number of ways by the number of ways to rotate the cube. So we have $3$ ways to place the reds. That is correct. So there are $3$ possible cubes.


A cube is painted so that one side is blue, two sides are red, and three sides are green. How many different such cubes can be painted? Two cubes are considered the same if one cube can be rotated in any way to match the second cube.