1. **Problem Statement:** Find the number of distinct arrangements of the letters in the word ABRACADABRA. 2. **Understanding the problem:** The word ABRACADABRA has 11 letters with some letters repeating. To find the number of distinct arrangements, we use the formula for permutations of multiset: $$\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}$$ where $n$ is the total number of letters, and $n_1, n_2, ..., n_k$ are the frequencies of each distinct letter. 3. **Count the letters:** - A appears 5 times - B appears 2 times - R appears 2 times - C appears 1 time - D appears 1 time 4. **Apply the formula:** $$n = 11$$ $$n_A = 5, n_B = 2, n_R = 2, n_C = 1, n_D = 1$$ $$\text{Number of arrangements} = \frac{11!}{5! \times 2! \times 2! \times 1! \times 1!}$$ 5. **Calculate factorials:** - $11! = 39916800$ - $5! = 120$ - $2! = 2$ 6. **Simplify denominator:** $$5! \times 2! \times 2! = 120 \times 2 \times 2 = 480$$ 7. **Final calculation:** $$\frac{39916800}{480} = 83160$$ **Answer:** There are **83160** distinct arrangements of the letters in ABRACADABRA.