Anagrams of the word "Abracadabra"

Question

Without any clear-shot way of verifying combinatorics problems, I'm not sure if my answer's correct or not. Is there a way to check? (Other than... brute force?)

Question: How many anagrams of the 11-letter word ABRACADABRA have no adjacent letter 'A's?

(Attempted) answer:

In our multiset, we have 5 A's, 2 B's, 2 R's, and a C and a D. We can permute the 2 B's, 2 R's, C and D in $\frac{6!}{2!2!}$ because we have repeated B's and R's. Each permutation of the 6 letters leave: 5 between the numbers and two on each end. That means we have ${7 \choose 5}$ ways to place the A's, so we have a total of $\frac{7!6!}{5!2!2!2!} = \frac{3}{4}7! = 3780$ ways.

Answer 1

$\color{green}\checkmark$ Yes, your reasoning is correct.

There are $\tbinom 6{2,2,1,1}$ ways to select places for two B, two R, one C, and one D in a string of six letters.

There are $\tbinom 75$ ways to select five from the seven spaces between (and beside) those letters in that string, to put the A, thus ensuring no A are adjacent to another.

Thus the answer is indeed $3780$: $$\dbinom 6{2,2,1,1}\dbinom 75$$