5. (COL) A cube is assembled with 27 white cubes. The larger cube is then painted black on the outside and disassembled. A blind man reassembles it. What is the probability that the cube is now completely black on the outside? Give an approximation of the size of your answer.
5. There are four types of small cubes upon disassembling:
(1) 8 cubes with three faces, painted black, at one corner:
(2) 12 cubes with two black faces, both at one edge;
(3) 6 cubes with one black face:
(4) 1 completely white cube.
All cubes of type (1) must go to corners, and be placed in a correct way (one of three): for this step we have $3^{5}$ - 8! possibilities. Further, all cubes of type (2) must go in a correct way (one of two) to edges, admitting $2^{12} \cdot 12$ ! possibilities; similarly, there are $4^{6} \cdot 6$ ! ways for cubes of type (3), and 24 ways for the cube of type (4). Thus the total number of good reassemblings is $3^{8} 8!\cdot 2^{12} 12!\cdot 4^{6} 6!\cdot 24$, while the number of all possible reassemblings is $24^{27}-27$ !. The desired probability is $\frac{3^{\star} 8!\cdot 2^{12} 121 \cdot 4^{0} 61: 24}{24^{27} \cdot 27!}$. It is not necessary to calculate these numbers to find out that the blind man practically has no chance to reassemble the cube in a right way: in fact, the probability is of order $1.8 \cdot 10^{-37}$.