You are given a cube. You are told to fill in each face randomly with some of the numbers $4, 5, 6, ..., 11$, with no repetition. What is the probability that for each two faces that are connected by a common edge, he two numbers written on them are co-prime?

Source: a mistaken understanding of that other question: Fill in numbers on the cube!

  • $\begingroup$ Wouldn't the answer depend on what numbers you choose to write? $\endgroup$ – msh210 Jul 6 '20 at 21:40
  • $\begingroup$ They are chosen randomly. $\endgroup$ – Florian F Jul 7 '20 at 6:59
  • $\begingroup$ "You are told to fill in each face with some of the numbers" does not sound like they're chosen randomly! I really recommend you edit the question to clarify. $\endgroup$ – msh210 Jul 7 '20 at 13:21

As with the other question, we first find out how many valid ways there are to place the numbers.

There are only four odd numbers, so we need at least two faces with an even number. They can't be adjacent so there are only two even numbers on the cube and they must be on opposite faces.
The opposite even faces are adjacent to all four odd-numbered faces, so cannot be 6 (shares a factor with 9) or 10 (shares a factor with 5). Therefore the even faces are 4 and 8.
Rotate the cube so that the 4 is on top, and the 5 in front. The bottom face must be the 8, and the other three odd numbers are coprime so can go in any order on the remaining faces. Up to rotation this gives $3!=6$ essentially different arrangements. With rotations this gives $6\cdot24=144$ valid number arrangements.

Now to get the probability:

We need to find the total number of possible arrangements. First choose 6 of the 8 numbers, and then arrange them on the faces in any order. That gives $\binom{8}{6}\cdot6!=20160$ ways.
The probability then becomes $\frac{144}{20160}=\frac{1}{140}$.

  • $\begingroup$ I got the same answer by other means. $\endgroup$ – Florian F Jul 7 '20 at 7:02

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