4. Elisa makes so-called fake dice. On each side of a fake die, one of the numbers 1 through 6 is written, but not every number has to appear, and some numbers may appear more often. However, from all sides, it must look like a real die. This means: at each corner, three different numbers come together, and no two of them add up to 7 (on a real die, such pairs of numbers are always opposite each other). The numbers 1, 2, and 4 can come together at a corner, for example, but 1, 2, and 5 cannot because $2+5=7$. A regular die, of course, also counts as a fake die.

Elisa is interested in the eye sum of the fake die: the sum of the six numbers on the fake die.

How many different possibilities are there for the eye sum of a fake die?
B4. 19 At each vertex, three different numbers always come together. Added together, these therefore amount to at least $1+2+3=6$. If we look at two opposite vertices, we count each face once and find that the eyesum is at least $2 \times 6=12$. In the same way, we can reason that the eyesum is at most $4+5+6+4+5+6=30$.

We will now show that all numbers from 12 to 30 can indeed occur as the eyesum of a pseudo-die. Consider the three pairs of opposite faces. For each of the faces in the first pair, we choose 3 or 4 as the number. For each of the faces in the second pair, we choose 2 or 5, and for each face in the third pair, we choose 1 or 6. In each vertex, there is now exactly one number from each of the three pairs $\{3,4\}$ and $\{2,5\}$ and $\{1,6\}$. The result is thus always a pseudo-die.

If you add the numbers in the first pair, the possible outcomes are therefore 6, 7, 8. The possibilities for the second pair are 4, 7, 10, and the possibilities for the third pair are $2, 7, 12$. If we add the four numbers from the first two pairs, the possible outcomes are thus $4+6, 4+7, 4+8, 7+6, 7+7, 7+8, 10+6, 10+7, 10+8$, or the numbers 10 to 18. The possibilities for the total eyesum are therefore the numbers $12 \mathrm{t} / \mathrm{m} 20, 17 \mathrm{t} / \mathrm{m} 25$ and $22 \mathrm{t} / \mathrm{m} 30$. Each number from 12 to 30 is thus the eyesum of a pseudo-die. In total, there are therefore 19 possible eyesums.