How many arrangements are possible in which no two $B$'s are next to each other?
I am approaching this problem by considering the complement.
Case 1: $BB$ is one letter. This leaves us with $1B, 2U, 2L, 1E, 2I, 1C, 1O, 1S $($12$ letters total)
Step 1: Place $BB$: $12$ ways
Step 2: Place the third $B$ such that it is not consecutive with $BB$. Is this step necessary? This would be split into two cases itself. One where BB is on the ends and one where BB is in the middle
Case 2: $BBB$ is one letter.
FINAL ATTEMPT:
The total number of arrangements of the $13$ letters in BUBBLELICIOUS is $\displaystyle\frac{13!}{3!(2!)^3}$.
Consider the complement. Consider $BB$ as one letter. Then there are $\displaystyle\frac{12!}{(2!)^3}$ ways to arrange the letters.
So there are $\displaystyle\frac{13!}{3!(2!)^3}-\displaystyle\frac{12!}{(2!)^3}$ arrangements where $BB$ is considered one letter.
However, we have counted twice where $BBB$ occurs. So we need to add this case back once.
Consider $BBB$ as one letter. The total number of arrangements in this case is $\displaystyle\frac{11!}{(2!)^3}$.
So the total number of arrangements in which no two $B$'s are next to each other is:
$$\displaystyle\frac{13!}{3!(2!)^3}-\displaystyle\frac{12!}{(2!)^3}+\displaystyle\frac{11!}{(2!)^3}$$
Link to part (b) of problem: Arrangements of BUBBLELICIOUS with no two consonants next to each other
I asked a different question because of this. My strategy is to first count all ways of arranging the B's without any being adjacent. If you use the relationship in this thread, this is $\binom{13-3+1}{3}$. After placing the B's, there are 10 letters left, for $10!$ placements; however, we must remove double counting of I's, U's, and L's (divide by 2 for each one), giving the result $$\binom{11}{3}\frac{10!}{8}$$