A child paints the six faces of a cube with six different co

Fun question, but I'm not a fan that the answer choices (1) are arranged not in ascending order and (2) include an ambiguous "none," which I take to mean "none of the above" rather than meaning "zero."

How many total ways are there to paint the cube? Let's paint the front face red. There are 5 choices for what to paint the back face, then 4 choices for what to paint the top, 3 choices for what to paint the right side, 2 choices for what to paint the bottom, and 1 choice for what to paint the left side. That's 5!. BUT, let's say the top/right/bottom/left were OYGB, that's the same as YGBO, which is the same as GBOY, which is the same as BOYG. So, we need to divide that 5! by 4. That leaves us with 5*3*2 = 30 total possible ways to paint the cube.

Now we need to find the number of ways to "win." If red is the front face, we need blue to be one of the adjacent faces and pink to be adjacent to each of those. Once the blue is painted, the pink could go on either side of the blue, so there are 2 choices for the pink. The remaining three colors can be completed in 3! ways. 2 ways to choose where to put pink * 3! ways to fill in OYG = 2*3*2 = 12.

There are 12 ways to "win" out of 30 total possible ways. That's 2/5.

Answer choice E.