Example 1-22 Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}, A_{8}$ be eight comrades, divided into 4 groups, each consisting of two people. How many different schemes are there?
Method 1: When $A_{1}$ chooses a partner from the same group, there are 7 possible choices. Among the remaining six people, one person has only 5 possible choices for a partner. Among the remaining four people, one person has only 3 possible choices for a partner. The last two people will form a group with no choice. Therefore, the number of ways $N=7 \times 5 \times 3=105$.

Method 2
The total number of permutations of 8 people is $8!$.
If 8 people are divided into 4 groups:
$$
\{1,2\},\{3,4\},\{5,6\},\{7,8\}
$$
1 and 2, 3 and 4, 5 and 6, 7 and 8 can be swapped, which results in the same grouping, so the repetition factor is $2^{4}$. Similarly, the 4 groups $\{1,2\},\{3,4\},\{5,6\},\{7,8\}$ can be permuted in 4! ways, all representing the same grouping, with a repetition factor of $4!$. Therefore, the number of groupings $N$ is
$$
N=\frac{8!}{4!2^{4}}=\frac{8 \times 7 \times 6 \times 5}{16}=105
$$

Method 3: When dividing 8 people into 4 groups, the first group has $\binom{8}{2}$ choices. Among the remaining 6 people, the second group has $\binom{6}{2}$ choices, and the third group has $\binom{4}{2}$ choices. Therefore, according to the multiplication rule,
$$
N=\binom{8}{2} \cdot\binom{6}{2} \cdot\binom{4}{2} / 4!=8 \times 7 \times 6 \times 5 \times 4 \times 3 / 2^{3} \cdot 4!=105
$$

The division by 4! is because the order of the 4 groups does not matter.
The above discussion can be generalized to the number of ways to group $2n$ members, with each group consisting of 2 people.