We want to draw a digit on each face of two identical cubes so that by appropriately rotating the cubes and placing them next to each other, we can read any day of the month, including single-digit days with a leading zero for uniformity. How many ways can we choose the digits to be drawn on each cube? (The relative positions of the six digits on one cube are not considered in the counting.)

![](https://cdn.mathpix.com/cropped/2024_05_02_9b382d5b0c4e2d739941g-1.jpg?height=353&width=591&top_left_y=398&top_left_x=756)
The numbers 1, 2, ..., 31 need to be displayed. The digits 0, 1, and 2 must be written on both dice because these need to be combined with a different digit from 7, and the remaining digits do not fit on one die. Thus, there are only $3-3$ spaces left on each die, totaling 6 spaces, while there are still 7 different digits to be placed.

However, the problem is solvable by noticing that a 6, when rotated $180^{\circ}$ in the plane, reads as a 9 (and that there is no need to display 69 or 96).

Thus, the question is how to divide the digits $3, 4, 5, 6, 7$, and 8 onto the $3-3$ empty spaces. Considering that the smallest digit, 3, will be on one of the dice, but only on one, the next space on this die can be filled with

$$
\text{the 4, the 5, the 6, or the } 7
$$

(but not the 8), and the last space can be filled with one of the larger digits, respectively,

$$
4-, \quad 3-, \quad 2-, \quad \text{or} \quad 1-
$$

ways to complete the design of our first die. By placing the unchosen digits on the other die, the problem can be solved with $4+3+2+1=10$ different pairs of dice.

![](https://cdn.mathpix.com/cropped/2024_05_02_cc66b578195f75a5563eg-1.jpg?height=290&width=859&top_left_y=863&top_left_x=622)

Our diagram illustrates this reasoning: each "run" starting from the 3 represents one possible arrangement. (A large number of possibilities would not be representable this way.)

Note. The reasoning can also be completed as follows: the 3 remaining spaces on the die labeled I can be filled in $\binom{6}{3}=20$ ways, and the remaining digits can be written on the die labeled II in every case. However, this way we would get each arrangement twice - by swapping I and II - so there are only $20: 2=10$ different possibilities.