1
$\begingroup$

enter image description here

Attempt:

$$\frac{(2^6 + 6 \cdot 2^3 + 3 \cdot 2^4 + 8 \cdot 2^2 + 6 \cdot 2^3)}{24} = 10$$

  • 0
    @AnandJ where did that formula you use possibly come from? The correct tool for this problem is [Burnside's Lemma](https://en.wikipedia.org/wiki/Burnside's_lemma) which the OP successfully used. Not to mention, $2^6 = 64\neq 32=2^5$ – 2017-02-13

1 Answers 1

3

$10$ appears correct for distinct cubes allowing rotation mapping. There are:
- $2$ solid-colour cubes $(B^6,R^6)$,
- $2$ cubes with one face different $(B^5R, BR^5)$,
- $2$ (opposite-face and adjacent-face) options for each of the $4{+}2$ colourings $B^4R^2, B^2R^4$
- and $2$ similar options for the evenly-split colouring $B^3R^3$,
which overall gives $1+1+2+2+2+1+1 = 10$

The restriction to rotations (no reflections) makes no difference in this case.