Non-standard dice

One can always rotate the cube to place it's face $1$ at the bottom.Then,we can do any of:

• Place any number other than $6$ at the top face,so we have 4 choices,& for each of them we can circularly permute the numbers on other squares in $(4-1)!$ ways, thus giving total $4*3!$ ways.

• Place 6 on the top face, then arbitrarily choose a front face,place a number $n$ ,then place any one of the 2 numbers other than $(7-n)$ on the opposite back face. (Note that this is the only other option,if we place $(7-n)$ on back face,the other two faces will also have sum $7$ ) Now since reflections are considered distinct, the other $2$ faces can permute in $2$ ways, giving in all $2*2$ ways.

Thus there are $4*3! + 2*2 = 28$ ways in all.

First, we count the total number of ways to label the faces. Consider the face labelled 1. There are $5$ ways to choose the face opposite 1. Now, there are four labels left to label the four sides around the middle. There are $4!$ ways to do this, but since rotations are not considered distinct, we must divide by $4$ , for each orientation formed by rotation. This is equivalent to the number of ways to seat four people around a table. Therefore, there are $5\cdot 6=30$ ways to label total.

It is much easier to count the complement, and subtract, rather than go through casework.We will count the number of ways each pair of opposite sides can add to 7, and subtract from 30.

To help visualize, you can use a net:

Consider the face labelled 1. 1 must be opposite 6.

$\begin{aligned} &|1| \\ |\_|&|\_||\_||\_| \\ &|6| \end{aligned}$

2 can be any of the middle squares, and 5 must be opposite. The choice of where 2 is doesn't matter, since rotations are not distinct.

$\begin{aligned} &|1| \\ |2|&|\_||5||\_| \\ &|6| \end{aligned}$

Now, 3 can be to the left or right of 2. These are distinct since reflections are distinct. 4 is opposite 3.

$\begin{aligned} &|1| \\ |2|&|3||5||4| \\ &|6| \end{aligned}$

Therefore, the answer is $30-2=\boxed{28}$ .

we have 720 permutations of 6 numbers 48 of which have 3 pairs that sum to 7 so 672 but rotations result in 24 counting of a setting so $\frac{672}{24}$ =28

Orient the die such that $1$ is at the "top" face. Then, either opposite $1$ is $6$ or opposite $1$ is not $6$ .

If $6$ is opposite $1$ , then on the remaining four faces, $2$ must not be opposite $5$ (this condition is sufficient, since in this case $3$ is opposite $4$ iff $2$ is opposite $5$ ). Thus, the middle four faces can read $2345, 2543, 2435, 2534$ . When constructing these faces, it is necessary that two paths don't share the same cycle, i.e. $2345 = 3452 = 4523 = 5234$ . Additionally, note that it doesn't matter if $1$ or $6$ is on top because replacing one with the other will simply result in a rotation of one of the orientations above.

If $6$ is not opposite $1$ , there are four choices of what number to put opposite. Then, the other $4$ faces can be oriented in any way. To make sure that one orientation is not a rotation of another, we must again ensure that each "cycle" is unique. Put four coordinates on a plane and label them $A, B, C, D$ . Since the path is cyclic we can choose any starting point. Then there are $3$ points that one could go to next, and after that there are another $2$ points, and then we must visit the last point and return to the beginning point, giving us $3! = 6$ cycles (this holds for all cycles of size $n$ ). Thus there are $4 * 6 = 24$ orientations in this case.

$24 + 4 = 28$

Without loss of generality, let the base number be 1 Suppose we ignore the constraints, how many ways are there to place the numbers? 5 ways to choose the number at the top. For the numbers at the sides, circular permutation of 4 numbers = 3!=6 5*6=30 Now consider the constraints. Number at the top must be 6 Then, we have to permutate 2 pairs (2,5) (3,4) in a circle so that the numbers are opposite each other. 2 ways to do that. Hence 30-2=28

We do this with complementary counting. First count the number of ways we can number a cube in total. Since rotations do not matter, we put the side with the number 6 on the face on the table. We then have a choice of 5 numbers to put on the top face.

The remaining 4 numbers go on the sides, but since we can rotate the 4 faces, its the same problem as arranging 4 numbers in a circle and we have $3!$ ways of arranging them. So there are $5\times3!=30$ ways of arranging the numbers on the faces.

Next we try and find the number of ways in which the opposite side do add up to 7. The only ways are the pairs $(1,6), (2,5), (3,4)$ . Once again, put 6 on the bottom and 1 must go on the top. Without loss of the number of permutations you can pick a pair of faces for the $(2,5)$ pair. Then we have 2 choices for the $(3,4)$ pair (3 goes either to the left or the right of 2).

So the required number of ways is $30-2=\fbox{28}$

Let's first count the number of ways of having the sum equal to 7; the opposite faces must be $\{3,4\},\{2,5\},\{1,6\}$ . Once we label a number on one side, the opposite side automatically becomes labeled; therefore, this yields $3!=6$ . However, we can always switch the two numbers on the opposite sides, so there are $2^3=8$ ways to do that. Therefore there are $6 \times 8=48$ ways of getting a sum of 7 when we ignore rotations and reflections.

There are a total of $6!$ ways of labeling the cube when we ignore rotations and reflections. To accommodate for this, we must divide by $6 \times 4=24$ .

Therefore, there are $\frac{6!-48}{24}=\boxed{28}$ ways.