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Question Number 66 by newuser last updated on 15/Nov/14

$$\mathrm{How}\mathrm{many}\mathrm{unique}\mathrm{arrangements}\mathrm{are}$$$$\mathrm{possible}\mathrm{for}\mathrm{a}2\times 2\mathrm{Rubic}\mathrm{cube}?$$

Commented by 123456 last updated on 14/Dec/14

$$6\times 5=30$$

Commented by 123456 last updated on 16/Dec/14

$$1+1+2+1=5$$

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