Key Concepts and Formulas

• Combinations: The number of ways to choose $k$ objects from a set of $n$ distinct objects, where order doesn't matter, is given by ${}^n{C_k} = \binom{n}{k} = \frac{n!}{k!(n-k)!}$.
• Permutations: The number of ways to arrange $n$ distinct objects in a row is $n!$.
• Multiplication Principle: If one event can occur in $m$ ways and another event can occur in $n$ ways, then the number of ways both events can occur is $m \times n$.

Step 1: Selecting the Novels

We need to select 4 novels out of 6 different novels. The order of selection doesn't matter, so we use combinations.

Number of ways to select 4 novels from 6 is: ${}^6{C_4} = \frac{6!}{4!(6-4)!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$ Explanation: We use combinations because choosing novels A, B, C, and D is the same as choosing D, C, B, and A. The order of selection does not matter.

Step 2: Selecting the Dictionary

We need to select 1 dictionary out of 3 different dictionaries. The order of selection doesn't matter, so we use combinations.

Number of ways to select 1 dictionary from 3 is: ${}^3{C_1} = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = \frac{3}{1} = 3$ Explanation: Since we are only selecting one dictionary, the number of ways to select it is simply the number of dictionaries available.

Step 3: Total Number of Ways to Select the Books

To find the total number of ways to select 4 novels and 1 dictionary, we multiply the number of ways to select the novels and the number of ways to select the dictionary, using the multiplication principle.

Total number of ways to select books = (ways to select novels) $\times$ (ways to select dictionary) $= {}^6{C_4} \times {}^3{C_1} = 15 \times 3 = 45$ Explanation: For each of the 15 ways to choose the novels, there are 3 ways to choose the dictionary.

Step 4: Arranging the Selected Books

We have selected 4 novels and 1 dictionary. The dictionary must be in the middle position. This leaves 4 positions for the 4 novels. The number of ways to arrange the 4 novels is $4!$.

Number of ways to arrange the 4 novels = $4! = 4 \times 3 \times 2 \times 1 = 24$ Explanation: Since the dictionary is fixed in the middle, we only need to arrange the 4 novels. There are 4 options for the first spot, 3 for the second, 2 for the third, and 1 for the last.

Step 5: Total Number of Arrangements

To find the total number of arrangements, we multiply the number of ways to select the books by the number of ways to arrange them.

Total number of arrangements = (ways to select books) $\times$ (ways to arrange books) $= 45 \times 24 = 1080$ Explanation: For each of the 45 ways to select the books, there are 24 ways to arrange them on the shelf with the dictionary in the middle.

Step 6: Comparing with Options

Now we compare our calculated value (1080) with the given options:

(A) at least 750 but less than 1000 (B) at least 1000 (C) less than 500 (D) at least 500 but less than 750

Our result, 1080, satisfies the condition "at least 1000".

Therefore, the correct option is (B).

Common Mistakes & Tips

• Carefully distinguish between combinations (selection) and permutations (arrangement). Order matters in permutations, but not in combinations.
• The constraint "dictionary in the middle" is crucial. It simplifies the arrangement step by fixing one position.
• Remember to use the multiplication principle when combining independent events.

Summary

We first selected 4 novels out of 6 and 1 dictionary out of 3, which gave us $15 \times 3 = 45$ ways to select the books. Then, we arranged these 5 books with the dictionary in the middle, which gave us $4! = 24$ ways to arrange them. Finally, we multiplied these two results to get the total number of arrangements: $45 \times 24 = 1080$. This corresponds to option (B).

The final answer is \boxed{1080}, which corresponds to option (B).