A kindergarten group consisting of 5 girls and 7 boys is playing "house." They select from among themselves a bride and groom, a mother, two flower girls, a best man for the bride and one for the groom, a bridesmaid for the bride and one for the groom. We know that three of the girls each have a brother in the group and there are no other sibling pairs. How many ways can the selection be made if siblings cannot be the bride and groom and the mother cannot be a sibling of either the bride or the groom? (Flower girls can only be girls, best men can only be boys, and there are no gender restrictions for the witnesses.)
Solution. We categorize the cases according to three criteria:

1. Does the bride have a sibling?
2. Does the groom have a sibling?
3. Is the family register holder a boy?

We will tabulate the possibilities.

| Case description | bride | groom | family register holder | bouquet girl | bridesmaid 1. | bridesmaid 2. | witness 1. | witness 2. | number of cases |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Both the bride and the groom have siblings, and the family register holder is a boy. | 3 | 2 | 5 | $\binom{4}{2}$ | 5 | 4 | 5 | 4 | 72000 |
| Both the bride and the groom have siblings, and the family register holder is a girl. | 3 | 2 | 3 | $\binom{3}{2}$ | 6 | 5 | 5 | 4 | 32400 |
| The bride has a sibling, but the groom does not, and the family register holder is a boy. | 3 | 4 | 5 | $\binom{4}{2}$ | 5 | 4 | 5 | 4 | 144000 |
| The bride has a sibling, but the groom does not, and the family register holder is a girl. | 3 | 4 | 4 | $\binom{3}{2}$ | 6 | 5 | 5 | 4 | 86400 |
| The bride does not have a sibling, but the groom does, and the family register holder is a boy. | 2 | 3 | 6 | $\binom{4}{2}$ | 5 | 4 | 5 | 4 | 86400 |
| The bride does not have a sibling, but the groom does, and the family register holder is a girl. | 2 | 3 | 3 | $\binom{3}{2}$ | 6 | 5 | 5 | 4 | 32400 |
| Neither the bride nor the groom has a sibling, and the family register holder is a boy. | 2 | 4 | 6 | $\binom{4}{2}$ | 5 | 4 | 5 | 4 | 115200 |
| Neither the bride nor the groom has a sibling, and the family register holder is a girl. | 2 | 4 | 4 | $\binom{3}{2}$ | 6 | 5 | 5 | 4 | 57600 |
| Total: |  |  |  |  |  |  |  |  | 626400 |

Therefore, the wedding can be realized in 626400 different ways as described in the problem.