1. **Problem statement:** We have 5 men and 4 women to be seated in a row such that no two women sit together. We need to find the number of ways to arrange them under this condition. 2. **Key idea:** To ensure no two women sit together, we can first arrange the men and then place the women in the gaps between men. 3. **Step 1: Arrange the men.** There are 5 men, so they can be arranged in $$5!$$ ways. 4. **Step 2: Identify gaps for women.** When 5 men are seated in a row, there are 6 possible gaps where women can be placed: one before the first man, four between the men, and one after the last man. 5. **Step 3: Place the 4 women in these 6 gaps.** Since no two women can sit together, each woman must occupy a separate gap. We need to choose 4 gaps out of 6 to place the women. The number of ways to choose these gaps is $$\binom{6}{4}$$. 6. **Step 4: Arrange the women in the chosen gaps.** The 4 women can be arranged among themselves in $$4!$$ ways. 7. **Step 5: Calculate total arrangements.** Multiply the number of ways to arrange men, choose gaps, and arrange women: $$\text{Total ways} = 5! \times \binom{6}{4} \times 4!$$ 8. **Step 6: Compute the values:** $$5! = 120$$ $$\binom{6}{4} = \frac{6!}{4! \times 2!} = \frac{720}{24 \times 2} = 15$$ $$4! = 24$$ 9. **Step 7: Final calculation:** $$120 \times 15 \times 24 = 43200$$ **Answer:** There are $$43200$$ ways to arrange 5 men and 4 women in a row such that no two women sit together.