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I was wondering how many possible configurations there are of a Rubik's cube of size greater than $3\times3\times3$ (e.g. for $4\times4\times4$, $5\times5\times5$, $\dots 10\times10\times10)$? We know that for a $3\times3\times3$-cube there are about $4.3 \times 10^{19}$ configurations, what about the bigger cubes?

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    $\begingroup$ If one follow the arguments in Jaap's puzzle page in the enumeration of $N_n$, the number of combinations of a $n \times n \times n$ Rubik's cube for $2 \le n \le 7$, one find $$N_n = \begin{cases} \frac{1}{24} \times 8!\cdot 3^7 \times 24!^{k-1}\times (24!/4!^6)^{(k-1)^2},& n = 2k\\ \\ 8!\cdot 3^7 \times 12!\cdot 2^{10} \times 24!^{k-1}\times (24!/4!^6)^{k(k-1)},& n = 2k+1 \end{cases} $$ In particular, for $n = 10$, we have $$N_{10} = \frac{8!\cdot 3^7 \times 24!^{20}}{24 \times 4!^{96}} \approx 8.298359851278236 \times 10^{349}$$ $\endgroup$ – achille hui Aug 2 '15 at 21:54
  • $\begingroup$ See also the OEIS page for this sequence: oeis.org/A075152. $\endgroup$ – Tad Aug 3 '15 at 5:02
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I'll give an upper bound.

We are considering a 10 by 10 cube here. A cube has 6 faces. Assuming that the cube has six colors, there should six options for each square on a face. There are a 100 squares per face. That means there are $6^{100}$ permutations per face. Multiplied by the six faces of the cube gives an upper bound of $6^{600}$ possibilities.

In an attempt to lower the bound, assume that after the first face, there are only 5 options left for each remaining square. After forming the permutations on the second face there would only be 4 options, etc. multiplying that out we get, $(720)^{100}$ as an estimate

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  • $\begingroup$ I just want to comment that my guess agrees with everything that achille hui posted... $\endgroup$ – Zach466920 Aug 5 '15 at 15:07

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