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This problem asks for the number of ways to distribute 12 distinct books into 3 distinct boxes, with the constraint that no box contains fewer than 3 books or more than 5 books.

### Step 1: Identifying Possible Distributions

We need to distribute 12 books such that each box contains at least 3 and no more than 5 books. The possible valid partitions of 12 into three parts (for the number of books in each box) that satisfy this condition are:

- \( (3, 4, 5) \)
- \( (4, 4, 4) \)

### Step 2: Counting the Number of Ways for Each Distribution

For each of the valid distributions, we need to calculate the number of ways to assign the distinct books to the boxes.

1. **Distribution (3, 4, 5):**
   - Choose 3 books for the first box from 12: \( \binom{12}{3} \).
   - Choose 4 books for the second box from the remaining 9: \( \binom{9}{4} \).
   - The remaining 5 books go into the third box.
   
   The number of ways for this distribution is:
   \[
   \binom{12}{3} \times \binom{9}{4} \times \binom{5}{5} = \frac{12!}{3! \times 4! \times 5!}
   \]

   But, since the boxes are distinct, the same arrangement can occur with the boxes in different orders. There are 3! ways to permute the boxes, so we divide by 3! to avoid overcounting. Thus, the total number of ways for this case is:
   \[
   \frac{12!}{3! \times 4! \times 5! \times 3!}
   \]

2. **Distribution (4, 4, 4):**
   - This case is simpler: divide the 12 books into three groups of 4.
   
   The number of ways to do this is given by:
   \[
   \binom{12}{4} \times \binom{8}{4} \times \binom{4}{4}
   \]
   
   Again, since the boxes are distinct, we divide by 3! to avoid overcounting:
   \[
   \frac{12!}{(4!)^3 \times 3!}
   \]

### Step 3: Calculate the Total Number of Ways

Now, we sum the number of ways from both cases. Let me calculate this.The total number of ways to pack 12 distinct books into 3 distinct boxes, such that no box contains fewer than 3 or more than 5 books, is **10,395**.

Would you like more details on any part of the solution, or do you have any further questions?

Here are 5 related questions you might find interesting:

1. What if the books were indistinct? How would that change the solution?
2. How many ways can the books be packed if there were 4 distinct boxes instead of 3?
3. How would the solution change if the boxes were indistinguishable (i.e., only the distribution matters, not the labeling of the boxes)?
4. What is the probability of randomly selecting one of these valid distributions from all possible distributions?
5. How would you approach this problem if the number of books was different, say 15?

**Tip:** When solving combinatorics problems, organizing possible cases and systematically applying counting principles can simplify even complex problems!

In how many ways can 12 distinct books be packed in three distinct boxes such that no box contains less than 3 or more than 5 books?

Learn how to calculate the number of ways to pack 12 distinct books into 3 distinct boxes, where each box contains between 3 and 5 books. This problem involves combinatorics, binomial coefficients, and permutations. Suitable for students in Grades 10-12.

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