4. Now we have keys numbered $1,2,3 \cdots \cdots 8$ and boxes numbered $1,2,3 \cdots \cdots 8$. Each key can only open the box with the same number. Now all the keys are placed into these boxes and locked, with each box containing only one key. How many different ways are there to place the keys so that at least two boxes need to be opened to unlock all the boxes? (Assume no keys are duplicated, and to open a box, either the corresponding key must be used or the box must be forced open, with no other methods)( )
A. 255
B. 20160
C. 14833
D. 35280
4. D

First, find the total number of ways to open only 1 box.
Arrange the numbers of the boxes to be opened in sequence, representing the order of opening, denoted as $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6} a_{7} a_{8}$. Then the key in box $a_{8}$ must be for box $a_{1}$, and the key in box $a_{i} (i=1,2 \cdots 7)$ must be for box $a_{i+1}$. In this case, if we do not start from box $a_{1}$ but from any one of the boxes $a_{i} (i=2,3 \cdots 8)$, only one box needs to be opened. Therefore, the number of ways to open only one box corresponds one-to-one with the circular permutations of $1,2 \cdots 8$.
Thus, there are 7! ways to place the keys such that only one box needs to be opened.
Therefore, the number of ways that meet the condition is $P_{8}^{8}-7!=35280$.