If you paint the six sides of a 3x3x3 cube and then slice it into 27 unit cubes, how many will have paint on 0, 1, 2, 3 of their square sides?
Generalize to any dimension:
If you paint the sides of a n dimensional cube with sides length 3 and slice it into 3n unit cubes, how many will have paint on 0, 1, 2, ..., n of their (n-1) dimensional sides?
0 sides = interior cubes = (n-2)^3
1 side = 6 sides * (n-2)^2 per side = 6(n-2)^2
2 sides = 12 edges *(n-2) per edge = 12(n-2)
3 sides = corners = 8
Checking, we see that this adds correctly:
n^3 -6n^2 + 12n + 8
6n^2 - 24n +24
12n - 24
8
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n^3
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oops. Misread the problem. I obviously answered for a 3 dimensional cube of side n.
Edited on March 5, 2015, 3:03 pm
Well this seems pretty straightforward.
