Answer: 1260

## Explanation This is a **multinomial coefficient** problem where we need to arrange 9 stocks into 3 distinct categories with specific counts: - 4 small-cap stocks - 3 blue-chip stocks - 2 emerging market stocks ### Step 1: Total arrangements without considering identical items If all 9 stocks were distinct, the total number of ways to arrange them would be: $$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880$$ ### Step 2: Accounting for identical items within categories Within each category, the stocks are considered identical for labeling purposes. We need to divide by the factorial of the number of items in each category to avoid overcounting: $$\text{Number of ways} = \frac{9!}{4! \times 3! \times 2!}$$ ### Step 3: Calculation $$4! = 24$$ $$3! = 6$$ $$2! = 2$$ $$\text{Number of ways} = \frac{362,880}{24 \times 6 \times 2} = \frac{362,880}{288} = 1,260$$ ### Step 4: Interpretation This represents the number of distinct sequences/labelings of the 9 stocks where we have exactly 4 small-cap, 3 blue-chip, and 2 emerging market stocks, and stocks within the same category are considered identical. **Therefore, the correct answer is 1,260 (Option A).**

Author: Nikitesh Somanthe