From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is:

We have 6 different novels and 3 different dictionaries in total. We must finally place exactly 5 books on the shelf - 4 of them novels and 1 of them a dictionary - and the special condition is that the dictionary must occupy the middle position of the row.

Because the middle (third) position is reserved for a dictionary, we first decide which dictionary will sit there. The rule for selecting one object out of many is the basic combination rule:

$${}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}$$

Here $$n=3$$ (three different dictionaries) and $$r=1$$. Therefore

$${}^{3}C_{1}=\frac{3!}{1!\,(3-1)!}=\frac{3\cdot2\cdot1}{1\cdot2\cdot1}=3.$$

So, there are 3 possible choices for the dictionary that will go in the middle.

Now we turn to the novels. We must pick 4 novels out of the 6 available. We again use the same combination formula with $$n=6$$ and $$r=4$$:

$${}^{6}C_{4}=\frac{6!}{4!\,(6-4)!}=\frac{6\cdot5\cdot4\cdot3\cdot2\cdot1}{(4\cdot3\cdot2\cdot1)\,(2\cdot1)}=\frac{720}{24\cdot2}=\frac{720}{48}=15.$$

Thus, there are 15 different sets of 4 novels that we might place beside the chosen dictionary.

After selecting the dictionary and the 4 novels, we must arrange the 5 books on the shelf. The dictionary’s position is already fixed in the middle, so only the 4 novels have freedom to be permuted among the remaining 4 slots (first, second, fourth, and fifth places). The number of ways to order 4 distinct items is the factorial of 4:

$$4!=4\cdot3\cdot2\cdot1=24.$$

Finally, by the fundamental principle of counting (multiplying the independent choices), the total number of admissible arrangements is

$$3\;(\text{choices of dictionary})\times 15\;(\text{choices of novels})\times 24\;(\text{permutations of those novels})=3\times15\times24.$$

We multiply step by step:

First, $$3\times15=45.$$

Next, $$45\times24=45\times(20+4)=45\times20+45\times4=900+180=1080.$$

So the exact count of valid arrangements is $$1080.$$

The problem asks only for a range. Since $$1080$$ is more than $$1000$$, it clearly falls in the category “At least 1000.”

Hence, the correct answer is Option B.