Geometry meets Combinatorics! (A cube problem)
A cube has each of its faces colored with a distinct color. How many distinct sets of colors can you "see" on the cube at a given time, given that you only have one eye?
Notes :
Consider your eye to be a fixed point in space outside the cube. We say that you can "see" a color if there exists a line which connects your eye to the center of the face which has that color which does not intersect with any other face of the cube.
The order in which the colors are seen is not important. {Red, Green, White} is the same as {Green, White, Red}
Hint/Extra Challenge : Prove, geometrically, that you can not "see" more than 3 colors at the same time.
The answer is 26.
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There are three cases.
First, let point E denote your eye and point C denote the center of the cube.
Case 1: When you can only see one color at a time.
This case occurs when the four faces adjacent to the plane which intersects with line E C are parallel to line E C . There are 6 C 1 = 6 possible ways to do this.
Case 2: When you can see two colors at a time.
This case occurs when line E C intersects with an edge, and the planes which are adjacent to the planes which share an edge are parallel to line E C . There are 6 C 2 = 1 5 ways to choose two faces at a time, but 3 ways are impossible as they choose non-adjacent faces, so there are 6 C 2 − 3 = 1 2 ways to do this.
Case 3: When you can see three colors at a time.
This case occurs when none of the above cases occur.
From the fact that you cannot see two non-adjacent faces at a time, you can only see one of the two paired non-adjacent faces. Now, if you choose three faces, you choose exactly one color from each pair. Since there are two ways you can choose from a pair, there are 2 ∗ 2 ∗ 2 = 8 ways you can choose one from three pairs.
So, our answer is 6 + 1 2 + 8 = 2 6 ways.