You have 500 identical red cubes and 500 identical black cubes.

How many ways can you arrange them in a $10\times10\times10$ "super-cube" such that no cube is in between two cubes of the same color (as each other) in any of the three Cartesian directions?

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Clarification:
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For any row of three cubes with abutting faces, the following combinations aren't allowed:

- RBR
- RRR
- BBB
- BRB

The answer is 70.

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Once you have defined a 2x2x2 cube in a corner, the rest of the cubes are determined by the following algorithm:

Let $l$ , $m$ , and $n$ represent the indices of the cubes in the $x$ , $y$ and $z$ direction.

$a_{l,m,n}$ given for $l > 3, m < 3, n < 3$ (8 values)

$a_{l+2,m,n}$ = The opposite color of $a_{l,m,n}$

$a_{l,m+2,n}$ = The opposite color of $a_{l,m,n}$

$a_{l,m,n+2}$ = The opposite color of $a_{l,m,n}$

And since there are 125 little 2x2x2 cubes this is an odd number, so in order to maintain an even ratio of black to red cubes, the first 2x2x2 cube must be picked with 4 red and 4 blacks.

Therefore the are $\binom{8}{4}$ ways to pick the colors in this first 2x2x2 cube.

Since the rest of the cube colors are determined from there, there are $\binom{8}{4} = \boxed{70}$ ways to choose the 1000 cubes.