Assume I have 4 “As” and 8 “B”s. How many strings of length 12 are there? How many strings of length 12 have no 2 “A”s adjacent?
I have been able to do the first part quite easily by finding 12!/4!8! but I am unable to do the next part - could anyone please help?
The number of arrangements of $4$ A's and $8$ B's with no two A's adjacent is the same as the number of arrangements of $4$ A's and $9$ B's with every A immediately followed by a B, which is the same as the number of arrangements of $4$ AB's and $5$ B's, so the answer is $\binom94=\frac{9!}{4!5!}=126$.