1.22 Divide a set containing 12 elements into 6 subsets, so that each subset contains exactly two elements. How many ways are there to do this?
[Solution 1] First, choose any two elements from 12 elements as one subset, there are $C_{12}^{2}$ ways to do this. Then, choose any two elements from the remaining 10 elements as one subset, there are $C_{10}^{2}$ ways to do this. Continue this process until the 12 elements are divided into 6 subsets, which results in $C_{12}^{2} \cdot C_{10}^{2} \cdot C_{8}^{2} \cdot C_{6}^{2} \cdot C_{4}^{2} \cdot C_{2}^{2}$ ways.

Since the 6 subsets are not concerned with the order, each way of dividing into 6 subsets is counted 6! times in the above counting process. Therefore, the number of ways that meet the requirement is
$$
S=\frac{1}{6!} C_{12}^{2} \cdot C_{10}^{2} \cdot C_{8}^{2} \cdot C_{6}^{2} \cdot C_{4}^{2}=11 \times 9 \times 7 \times 5 \times 3=10395 \text {. }
$$
[Solution 2] Label the 12 elements as $a_{1}, a_{2}, \cdots, a_{12}$. The first subset, besides containing $a_{1}$, also contains one more element, which can be chosen from the remaining 11 elements, giving 11 ways. Then, from the remaining 10 elements, choose the one with the smallest index and another element from the remaining 9 elements to form the second subset, giving 9 ways. Continue this process, and the resulting ways of division are all different, so the total number of ways is
$$
S=11 \times 9 \times 7 \times 5 \times 3=10395 .
$$