1. Select the books

We need:

• $4$ novels from $6$ different novels
• $1$ dictionary from $3$ different dictionaries

Number of ways to choose the novels: $\binom{6}{4} = 15$

Number of ways to choose the dictionary: $\binom{3}{1} = 3$

So, total ways to select the books: $15 \times 3 = 45$

2. Arrange them with the dictionary always in the middle

There are $5$ books in total. If the dictionary must always be in the middle, then its position is fixed at the $3^{\text{rd}}$ place.

The remaining $4$ selected novels can be arranged in the other $4$ positions in: $4! = 24$ ways.

3. Total number of arrangements

Hence, total number of required arrangements is: $\binom{6}{4} \cdot \binom{3}{1} \cdot 4!$ $= 15 \times 3 \times 24$ $= 1080$

4. Match with options

Since $1080 \ge 1000,$ it falls in:

Option C: at least 1000

5. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So, the answer agrees with the stored correct answer.