# Fill in numbers on the cube!

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

• Note that if it is an ongoing competition you are not supposed to post the problem here. Not before it has ended. Jul 3, 2020 at 14:33
• @FlorianF It is from 2019. Jul 3, 2020 at 23:35

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 $$\frac{96}{40320}=\frac{1}{420}$$.

• Correct! Checkmark incoming! Jul 4, 2020 at 12:18