You're right to question the multiplication, and it's important to clarify why it's done.

### Explanation:
The problem involves placing 9 balls into 9 boxes, with the restriction that 5 of the balls cannot be placed into 3 smaller boxes.

Here’s a step-by-step breakdown:

### Step 1: Place the 5 Restricted Balls
- **Restricted Balls**: We have 5 balls that cannot go into 3 smaller boxes. So, these 5 balls must go into the remaining 6 larger boxes.
- We can choose 5 out of the 6 larger boxes to place the restricted balls. The number of ways to choose 5 boxes from the 6 is \( \binom{6}{5} \). However, \( \binom{6}{5} \) is equal to 6 (since choosing 5 out of 6 is the same as choosing 1 out of 6 that will be empty).
- After selecting the 5 boxes, we can arrange the 5 balls in those boxes in \( 5! \) ways (since there is no repetition).

So, the number of ways to place the 5 restricted balls into the 6 larger boxes is:
\[
6 \times 5! = 6 \times 120 = 720
\]

### Step 2: Place the Remaining 4 Unrestricted Balls
- **Unrestricted Balls**: The remaining 4 balls can be placed freely into the 4 remaining boxes, which includes the 3 smaller boxes and 1 remaining larger box.
- These 4 balls can be arranged in \( 4! \) ways since they are being placed one per box.

Thus, the number of ways to place the 4 unrestricted balls into the remaining 4 boxes is:
\[
4! = 24
\]

### Step 3: Multiply the Two Results
- The reason we multiply \( 4! \) with 720 is because of the **rule of multiplication in combinatorics**: when two independent events occur in sequence, you multiply the number of ways each event can occur.
- In this case, placing the restricted 5 balls and the unrestricted 4 balls are two independent tasks. So, we multiply the number of ways to place the 5 restricted balls by the number of ways to place the 4 unrestricted balls.

Thus, the total number of ways is:
\[
720 \times 24 = 17,280
\]

### Conclusion:
We multiply \( 4! \) with 720 because we are considering two independent stages of the problem: 
1. Arranging the restricted 5 balls in the 6 larger boxes.
2. Arranging the remaining 4 unrestricted balls in the 4 remaining boxes.

This is a typical application of the multiplication rule in counting.

Let me know if you need more details! Here are 5 follow-up questions:
1. What would happen if 3 of the restricted balls couldn’t fit in 2 boxes instead of 5 balls and 3 boxes?
2. Can you explain when to use combinations \( \binom{n}{k} \) versus permutations \( P(n, k) \)?
3. What is the result if the restriction is removed entirely?
4. How does this problem change if we allow some boxes to remain empty?
5. How would the result change if there were repetitions allowed in placing the balls?

**Tip:** When multiple stages are involved in a combinatorial problem, always ensure to multiply the results of independent steps!

Explore the reasoning behind multiplying factorials in combinatorial counting problems. Learn step-by-step how to solve problems involving factorial multiplication, such as arranging balls into boxes with restrictions. Suitable for advanced high school students and enthusiasts of combinatorics.

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