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Question Number 66 by newuser last updated on 15/Nov/14
$$\mathrm{How}\:\mathrm{many}\:\mathrm{unique}\:\mathrm{arrangements}\:\mathrm{are}\: \\ $$$$\mathrm{possible}\:\mathrm{for}\:\mathrm{a}\:\mathrm{2}×\mathrm{2}\:\mathrm{Rubic}\:\mathrm{cube}? \\ $$
Commented by 123456 last updated on 14/Dec/14
$$\mathrm{6}×\mathrm{5}=\mathrm{30} \\ $$
Commented by 123456 last updated on 16/Dec/14
$$\mathrm{1}+\mathrm{1}+\mathrm{2}+\mathrm{1}=\mathrm{5} \\ $$