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

Source: HK Prelim 2019 Q20

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    $\begingroup$ Note that if it is an ongoing competition you are not supposed to post the problem here. Not before it has ended. $\endgroup$
    – Florian F
    Jul 3 '20 at 14:33
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    $\begingroup$ @FlorianF It is from 2019. $\endgroup$
    – Culver Kwan
    Jul 3 '20 at 23:35
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Let's first see in how many ways the numbers can be placed with all neighbours coprime.

The four even numbers must be non-adjacent, and the only way to have four non-adjacent vertices of a cube is when they form a regular tetrahedron. There is really only one way to arrange four numbers in a tetrahedron up to rotation and reflection.
Once you have the tetrahedron of 4 even numbers, the 9 cannot be adjacent to the 6, which leaves only the vertex of the cube diametrically opposite the 6. The same goes for the 5 and the 10.
That leaves the 7 and 11, which are coprime to everything. They can be placed either way in the last two spots.
That means that up to rotation and reflection, there are only 2 ways to arrange the numbers. The group of symmetries of the cube has size 48, so after rotations and reflections there are $2\cdot48=96$ valid arrangements.

Now for the probability:

There are $8!=40320$ ways to arrange the numbers, of which $96$ are valid. The probability is therefore $\frac{96}{40320}=\frac{1}{420}$.

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  • $\begingroup$ Correct! Checkmark incoming! $\endgroup$
    – Culver Kwan
    Jul 4 '20 at 12:18

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