1. **Problem statement:** We need to find the number of ways to arrange 5 different mathematics books and 4 different physics books on a shelf such that no two physics books are adjacent. 2. **Key idea:** To ensure no two physics books are together, we first arrange the mathematics books and then place the physics books in the gaps between them. 3. **Step 1: Arrange mathematics books.** There are 5 different mathematics books, so they can be arranged in $$5!$$ ways. 4. **Step 2: Identify gaps for physics books.** When 5 mathematics books are arranged, there are $$5 + 1 = 6$$ gaps (including the ends) where physics books can be placed. 5. **Step 3: Place physics books.** We have 4 physics books to place in these 6 gaps, with at most one physics book per gap to avoid adjacency. 6. **Step 4: Choose gaps for physics books.** Number of ways to choose 4 gaps out of 6 is $$\binom{6}{4}$$. 7. **Step 5: Arrange physics books in chosen gaps.** The 4 physics books can be arranged in $$4!$$ ways. 8. **Step 6: Calculate total arrangements.** Multiply all parts: $$\text{Total ways} = 5! \times \binom{6}{4} \times 4!$$ 9. **Step 7: Compute values:** $$5! = 120$$ $$\binom{6}{4} = \frac{6!}{4!2!} = 15$$ $$4! = 24$$ 10. **Final answer:** $$120 \times 15 \times 24 = 43200$$ So, there are **43200** ways to arrange the books so that no two physics books are together.