Number of ways of making a die using the digits $1,2,3,4,5,6$

Find the number of ways of making a die using the digits $1,2,3,4,5,6$.

I know that $6!$ is not the correct answer because some arrangements can be obtained just by rotation of the dice. So there will be many repetitions. I tried by fixing any two opposite faces and using circular permutations for the remaining 4 faces. Number of such arrangements is $2!\binom{6}{2}(4-1)!=2\times15\times 6=180$.

But answer given is just $30$. Maybe there are still some repetitions which I am not seeing.

• There's no restriction that opposite sides should sum up to 7? – peter.petrov Mar 14 '16 at 15:03
• @peter.petrov No restrictions – Aditya Dev Mar 14 '16 at 15:05
• You have overcounted in two ways. Your $2!$ flips the first two faces you picked, but those are equivalent. Your $6 \choose 2$ picks the first two faces, but you get the same die if you pick two others that would have been opposite among the final four. You should place the $6$ first, then pick the opposite face, which has $5$ choices. That $5$ compared to your $30$ is the excess factor $6$ – Ross Millikan Mar 14 '16 at 15:07

fix one number on one face, then choose any of the 5 remaining numbers to go on the opposite face. the remaining 4 numbers can be arranged in a ring in $3!=6$ distinguishable ways, hence the answer is 30

• Why is this different from my method? – Aditya Dev Mar 14 '16 at 15:04
• because you need to fix one number to one face. if you allow all 6 numbers to occupy in turn one particular face, you overcount by a factor of 6 – David Quinn Mar 14 '16 at 15:49

Here's yet another way of thinking: $6!$ arrangements divided by the $4!=24$ rotational symmetries of the cube, so there are $5 \cdot 6=30$ ways.

Fix $1$ (say) at bottom, and any of $5$ others at the top, in $5$ ways.

Now remember that you can rotate the die w/o disturbing the top and bottom to show $4$ different faces, so the remaining numbers can be fixed in $4!/4 = 6$ ways

$5 \times 6 = 30$

The faces on a die are analogous to the vertices on an octahedron. You are correct in beginning with $6!.$ Now, we see how many times we have overcounted. Notice that for a fixed top vertex of the octahedron, which can be labeled as any of the $6$ numbers, there are four rotational congruencies. And as I have just mentioned, any of the six faces could be positioned as the top vertex. So each arrangement was counted $6 \times 4 = 24$ times. Accounting for this, the number of arrangements is $\frac{6!}{24} = \frac{720}{24} = \boxed{30}.$

Hint if you fix an arrangement then it has $6$ similar arrangements you see why by rotating in $2$ directions X,Z so answer should be $\frac{180}{6}=30$

• Its difficult to imagine that. – Aditya Dev Mar 14 '16 at 15:08
• See hold a dice it can be rolled in front direction or sideways – Archis Welankar Mar 14 '16 at 15:09