1. **State the problem:** Find the number of distinguishable permutations of the letters in the string "AAABBBCD". 2. **Formula used:** The number of distinguishable permutations of a multiset of $n$ elements where there are groups of identical elements is given by: $$\frac{n!}{n_1! \times n_2! \times \cdots \times n_k!}$$ where $n$ is the total number of letters, and $n_1, n_2, \ldots, n_k$ are the counts of each distinct letter. 3. **Count the letters:** - A appears 3 times - B appears 3 times - C appears 1 time - D appears 1 time Total letters $n = 3 + 3 + 1 + 1 = 8$. 4. **Apply the formula:** $$\frac{8!}{3! \times 3! \times 1! \times 1!}$$ 5. **Calculate factorials:** $$8! = 40320$$ $$3! = 6$$ 6. **Simplify the expression:** $$\frac{40320}{6 \times 6} = \frac{40320}{36}$$ 7. **Perform the division:** $$\frac{40320}{36} = 1120$$ **Final answer:** There are **1120** distinguishable permutations of the letters in "AAABBBCD".