3. In how many ways can the set $\{1,2, \ldots, 12\}$ be divided into six disjoint two-element subsets such that each of them contains coprime numbers (i.e., numbers that do not have a common divisor greater than one)?
3. No two even numbers can be in the same of the six two-element subsets (let's call them "pairs"), and we can thus limit ourselves to such partitions of the set $\{1,2, \ldots, 12\}$ (let's call them even-odd), in which pairs are always composed of one even and one odd number. Another restriction is that the numbers 6 and 12 cannot be paired with the numbers 3 and 9; the number 10 cannot be paired with the number 5. We will determine the number of valid even-odd partitions in two ways.

FIRST WAY: First, we determine the potential even "partners" for each odd number from 1 to 11. For the numbers 1, 7, and 11, the set is $\{2,4,6,8,10,12\}$, for the numbers 3 and 9, the set is $\{2,4,8,10\}$, and finally, for the number 5, the set is $\{2,4,6,8,12\}$. From this, it is clear that we cannot directly apply the product rule to determine the number of all valid selections of even partners, regardless of the order in which we select for the given odd numbers. However, for the "promising" order $(5,3,9,1,7,11)$, the product rule works if we distinguish in advance whether the number 5 is paired with a number from $\{6,12\}$ or a number from $\{2,4,8\} .{ }^{1}$ There are thus $2 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 1=144$ possibilities in the first case and $3 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1=108$ in the second. In total, we have $144+108=252$ possibilities.

SECOND WAY: Since in an even-odd partition, each of the six odd numbers must be paired with one of the six even numbers, the number of all such (valid and invalid) partitions is equal to $6 !=720$. We will now count how many of these even-odd partitions are invalid.

All invalid even-odd partitions clearly form the set $M_{6} \cup M_{12} \cup M_{10}$, where $M_{6}$ is the set of even-odd partitions of the set $\{1,2, \ldots, 12\}$ in which the number 6 is paired with the number 3 or 9, $M_{12}$ is the set of even-odd partitions in which the number 12 is paired with the number 3 or 9, and finally, $M_{10}$ is the set of partitions in which the number 10 is paired with the number 5. The size of the union $M_{6} \cup M_{12} \cup M_{10}$ is, according to the principle of inclusion and exclusion, whose validity can be seen for three sets using a Venn diagram (Fig. 2), equal to

$$
\begin{aligned}
& \left|M_{6}\right|+\left|M_{12}\right|+\left|M_{10}\right|-\left|M_{6} \cap M_{12}\right|-\left|M_{6} \cap M_{10}\right|-\left|M_{12} \cap M_{10}\right|+\left|M_{6} \cap M_{12} \cap M_{10}\right| \\
& =2 \cdot 5 !+2 \cdot 5 !+5 !-2 \cdot 4 !-2 \cdot 4 !-2 \cdot 4 !+2 \cdot 3 !=468 .
\end{aligned}
$$

Since there are 720 even-odd partitions in total and 468 are invalid, the number of valid ones is $720-468=252$.

![](https://cdn.mathpix.com/cropped/2024_04_17_b6718dc1fa139fe24a99g-4.jpg?height=437&width=506&top_left_y=1809&top_left_x=775)

Fig. 2

Award 6 points for a complete solution. If the solver correctly determines the result in a "combinatorial form" (e.g., as $2 \cdot 4 \cdot 3 \cdot 3 \cdot 2 \cdot 1+3 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1$) and makes a mistake only in its calculation, award 5 points. If they present a functional method to arrive at the result (other than listing all possibilities) and do not complete the solution or make a mistake in the process, award no more than 3 points. Do not award points for observations about even-odd partitions or for listing coprime pairs. [^0]