【Example 1】Four people, A, B, C, and D, are practicing table tennis doubles. Two people form a team. How many different ways are there?
Here, four people need to be divided into two teams, meaning two groups. The numbering of the groups is irrelevant; the substantive issue is who is teamed up with whom? Clearly, there are 3 ways to group them: (1) Team A and B, Team C and D; (2) Team A and C, Team B and D; (3) Team A and D, Team B and C.

However, if calculated using the combination model, there are 6 ways to group them:

They are specifically:
(1) A and B as Group 1, C and D as Group 2;
(2) A and B as Group 2, C and D as Group 1;
(3) A and C as Group 1, B and D as Group 2;
(4) A and C as Group 2, B and D as Group 1;
(5) A and D as Group 1, B and C as Group 2;
(6) A and D as Group 2, B and C as Group 1.

From this, to derive the number of unnumbered groupings from the number of numbered groupings, one only needs to divide the former by the number of different ways to arrange the group numbers. Specifically, here it should be divided by \(2!\), i.e.,
$$
\frac{1}{2!}\binom{4}{2}=\frac{4!}{2!2!2!}=3
$$

But this assertion may not be comprehensive; another example can show this.