# Cubeception

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

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

The answer to the above question is $8$ and it is so because each corner cube is part of a unique $2\times 2$ selection.

Now, how many selections of $1 \times 1$ cubes can be made from a $3\times 3$? The answer here is too simple and is $27$.

And the number of selections of a $3\times 3$ cube from a $3\times 3$ is $1$.

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

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

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

Note by Raghav Vaidyanathan
5 years, 6 months ago

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

- 5 years, 6 months ago

Great explanation! Thanks!

- 5 years, 6 months ago