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.

Please post your ideas. Thanks!

#Combinatorics #Counting #RubiksCube #Interesting

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

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 $(n+m-1)^3$ .

Daniel Liu - 5 years, 11 months ago

Great explanation! Thanks!

Raghav Vaidyanathan - 5 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$