How many ways 6 girls and 3 boys are arranged in a circle so that no boys next to each other?

The title has said everything. Anyway, this is the original problem:

My attempt:

$A:=$ boys are next to each other.

$B :=$ boys aren't next to each other.

$$|A\cup B| = (9-1)!$$

$$|A| = (7-1)!\cdot3!$$

I thought it could be solved by inclusion-exclusion principle, but I couldn't think of $|A\cap B|$. However, I'm not sure if this is correct. I need to find $|B|$. Or is it just enough to solve this:

$$|A\cup B| - |A|$$

You attempted to solve a different problem from the one that was asked, namely the number of circular arrangements of six girls and three boys in which not all three boys are consecutive. However, we want to find the number of arrangements in which no two of the boys are adjacent.

Suppose one of the girls is Amy. Seat her. Relative to Amy, the remaining five girls may be seated in $5!$ ways as we proceed clockwise around the circle. This creates six spaces in which to place the boys, one to the left of each girl. To ensure that no two of the boys are adjacent, we must choose three of these six spaces in which to place a boy, which can be done in $\binom{6}{3}$ ways. The three boys can be arranged in the selected spaces in $3!$ ways. Hence, the number of circular seating arrangements of six girls and three boys in which no two boys are adjacent is $$5!\binom{6}{3}3!$$