1. **Problem Statement:** We need to find the number of ways to arrange four items from three different departments in a one-page advertisement with 3 rows and 4 columns, such that items from the same department are in the same row. 2. **Understanding the problem:** There are 3 rows and 4 columns, total 12 positions. We have 3 departments and 4 items to place. Items from the same department must be placed in the same row. 3. **Key points:** - Since there are 3 departments and 3 rows, each department's items must occupy exactly one row. - The total number of items is 4, so the distribution of items per department must sum to 4. - Each row can hold up to 4 items. 4. **Possible distributions of 4 items into 3 departments (rows):** - (2,1,1) items per department (since 3 departments and total 4 items) 5. **Step 1: Assign departments to rows:** - Number of ways to assign 3 departments to 3 rows is $3! = 6$. 6. **Step 2: Distribute 4 items into departments with counts (2,1,1):** - Choose which department has 2 items: 3 ways. - The other two departments have 1 item each. 7. **Step 3: Choose which items go to which department:** - Total 4 items, choose 2 for the department with 2 items: $\binom{4}{2} = 6$ ways. - Remaining 2 items go one each to the other two departments. 8. **Step 4: Arrange items within each row:** - For the row with 2 items: number of ways to arrange 2 items in 4 positions is $P(4,2) = \frac{4!}{(4-2)!} = 12$. - For each row with 1 item: number of ways to arrange 1 item in 4 positions is $P(4,1) = 4$. 9. **Step 5: Calculate total arrangements:** - Assign departments to rows: 6 ways - Choose department with 2 items: 3 ways - Choose items for that department: 6 ways - Arrange items in rows: $12 \times 4 \times 4 = 192$ ways 10. **Final total:** $$6 \times 3 \times 6 \times 192 = 20736$$ **Answer:** There are $20736$ possible arrangements.