Round Table permutation with more chairs than people

A group of six boys and three girls went for a dinner after the show. They were given a round table for twelve people.

a) Find the number of possible arrangements if seats are not numbered and the group can be seated without restriction.
b) Find the number of possible arrangements if seats are numbered and the three girls have to sit together as a group, with empty seats to separate them from the boys.

$${12 \choose 9} \cdot \frac{9!}{12}$$
1. The three girls have to sit next to each other, so select a chair to seat one of the girls, and put two other girls next to her. If the boys are expected to sit together as a group, then either assign one empty chair to the left and two to the right, or two to the left and one to the right. Then, for the remaining chairs, arrange the six boys. In total, the number of arrangements is: $${12 \choose 1} \cdot 3! \cdot 2 \cdot 6!$$ If the boys do not have to sit together as a group, assign one empty chair to the left and one to the right of the girls, and arrange the six boys among the seven remaining chairs. In total, the number of arrangements is: $${12 \choose 1} \cdot 3! \cdot 7!$$