Recently, I was playing with my two year old brother. There was an old 3×33\times 3 Rubik's cube lying around and my dad posed a simple problem:

How many selections of 2×22\times 2 cubes can be made from a 3×33\times 3 Rubik's cube?

The answer to the above question is 88 and it is so because each corner cube is part of a unique 2×22\times 2 selection.

Now, how many selections of 1×11 \times 1 cubes can be made from a 3×33\times 3? The answer here is too simple and is 2727.

And the number of selections of a 3×33\times 3 cube from a 3×33\times 3 is 11.

1,8,271, 8, 27.... Reminded me of something. This note is an attempt to prove or disprove the following hypothesis.

In an n×nn \times n Rubik's cube, the number of selections of solid cubes of edge length m(n)m(\le n) is (nm+1)3(n-m+1)^3

I am convinced that there is an elegant solution to this problem and that the hypothesis is true.

Please post your ideas. Thanks!

Note by Raghav Vaidyanathan
6 years ago

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Consider one dimension first. Given an array of length nn, we can place a smaller array of length mm in nm+1n-m+1 positions. This is the same for all three dimensions, bringing the total number of placings to be (n+m1)3(n+m-1)^3.

Daniel Liu - 6 years ago

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Great explanation! Thanks!

Raghav Vaidyanathan - 6 years ago

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