$11$ different books are on a $3$-shelf bookcase. In how many different ways can the books be arranged such that at most one shelf is empty?

$ \textbf{(A)}\ 75\cdot 11!
\qquad\textbf{(B)}\ 62\cdot 11!
\qquad\textbf{(C)}\ 68\cdot 12!
\qquad\textbf{(D)}\ 12\cdot 13!
\qquad\textbf{(E)}\ 6 \cdot 13!
$
1. **Understanding the problem**: We need to arrange 11 different books on a 3-shelf bookcase such that at most one shelf is empty. This means we can have either:
   - One shelf empty and the other two shelves with books.
   - No shelf empty (all three shelves have books).

2. **Total arrangements without any restriction**: If there were no restrictions, we could place the 11 books in any of the 3 shelves. Each book has 3 choices, so the total number of ways to place the books is:
   \[
   3^{11}
   \]

3. **Subtracting the invalid cases**: We need to subtract the cases where more than one shelf is empty. There are two such cases:
   - All books on one shelf.
   - Two shelves empty.

4. **All books on one shelf**: There are 3 ways to choose which shelf will contain all the books. The number of ways to arrange the books on that shelf is:
   \[
   3 \cdot 11!
   \]

5. **Two shelves empty**: This is not possible because we need at least 11 books on one shelf, which is already considered in the previous step.

6. **Valid arrangements**: The valid arrangements are:
   \[
   3^{11} - 3 \cdot 11!
   \]

7. **Simplifying the expression**: We need to simplify the expression \(3^{11} - 3 \cdot 11!\). However, the given solution uses a different approach by simulating the shelves with two indistinguishable book-ends.

8. **Using the book-ends approach**: We can simulate the shelves with two indistinguishable book-ends. There are \(\frac{13!}{2!}\) ways to arrange the books and book-ends. We subtract the cases where all books are on one shelf:
   \[
   \frac{13!}{2!} - 3 \cdot 11!
   \]

9. **Simplifying further**: 
   \[
   \frac{13!}{2!} - 3 \cdot 11! = \left(\frac{1}{2} \cdot 13 \cdot 12 - 3\right) \cdot 11!
   \]
   \[
   = \left(78 - 3\right) \cdot 11!
   \]
   \[
   = 75 \cdot 11!
   \]

Conclusion:
\[
\boxed{75 \cdot 11!}
\]