Number of ways to label a die

Question

I think I am on the right track on this probability question but just wanted to get it checked.

“If you have a regular die and you wipe the numbers off the sides of the die. How many different unique ways can you put the numbers (1-6) on the die? (So that if you turn the die, there is no same combination)”

I was thinking that this would simply be $6!$ (factorial). Figured that one side has $6$ numbers that could be picked. Once that is done, the next side only has $5$ numbers. Next $4$, then $3$ etc.

Answer 1

First, notice that there are $24$ ways of turning the die. This is because if you consider one particular face of the die, there are exactly $4$ ways of turning it so that this face is always in the same place. Then do the same for the other faces, and get $4 \times 6 = 24$.

$6!$ is the number of ways to number the die but this also counts the $24$ possible turns of the die. So the answer is $\frac{6!}{24}=30$.

Answer 2

Write the number $1$ on one of the faces. You then have a choice of $5$ numbers to go on the opposite face.

For each such choice, the remaining $4$ numbers can be arranged in $\frac{4!}{4}=6$ ways, and hence the total is $5\times 6=30$

Answer 3

There is famous recreational toy connected with this question: Mac Mahon's cubes.

Assume the die is standing on the face colored $6$. There are $5$ choices for the top face. The remaining $4$ colors can be paired off in $3$ ways to ornate opposite vertical faces, and the resulting pairs can be arranged in $2$ mirror equivalent ways on these faces. Makes $5\cdot3\cdot2=30$ colorings.