This Is Insane

Probability Level 2

Find the number of permutations of the non-supercube i.e $12 \times 12 \times 12$ Rubik's Cube . What ever the answer is multiply it by $\displaystyle \frac {(24)^{144}}{(24!)^{19}}$ and mark the option.

Take it as a challenge ! It is fun to solve.

You can use wolfram alpha for better calculation.

\displaystyle 1.0084... \times 10^{260}

\displaystyle 3.35847... \times 10^{260}

\displaystyle 2.41... \times 10^{260}

1 solution

Nikola Alfredi
Feb 24, 2020

SOLUTION

General formula for any $n \times n \times n$ Cube :

$\displaystyle \frac {(8! \times 3^7 ) ( \frac {1}{(4!)^6} 24! )^{\lfloor \frac {1}{4} (n-2)^2 \rfloor } (24!)^{\lfloor \frac {1}{2} (n-2) \rfloor } (12! \times 2^{11})^{n \mod 2}} {(24 - 23(n \mod 2) ) \times 2^{n \mod 2}}$

If you want to know more about how the formula came Click Here

If you plug in 12 instead of $n$ then Exact answer

$\displaystyle (8! \times 3^7 ) \times \frac {(24!)^{30}}{24^{151}} \approx 2.0636778980738443576466031979202588901702515160010261... × 10^{513}$

But the answer is not this !

ANSWER : $\displaystyle (8! \times 3^7 ) \times \frac {(24!)^{30}}{24^{151}} \times \frac {(24)^{144}}{(24!)^{19}} \approx 1.0084467862454126563877397595078962430925501867714016... × 10^{260}$

This Is Insane

1 solution

0 pending reports