*Three. (20 points) 4 small balls with a radius of $r$ are placed in cylinder $A$, numbered from top to bottom as $1,2,3,4 . A$'s base radius is slightly larger than $r . B, C$ are cylinders identical to $A$. The balls in $A$ are moved unidirectionally to $C$ via $B$, meaning no balls can be moved from $C$ to $B$, or from $B$ to $A . B$ can temporarily store several balls, but must follow the "last in, first out" rule. How many different arrangements of the balls are possible in $C$ after all the balls have been moved into $C$?
Three, first consider the case of three small balls. At this time, there are a total of 5 arrangement methods, each method (from bottom to top) and the operation process is represented as follows:
(1)(2)(3) (1) $\rightarrow B \rightarrow C$, (2) $\rightarrow B \rightarrow C$, (3) $\rightarrow B \rightarrow$ C)
(1) (3) (2) (1) $\rightarrow B \rightarrow C$, (2) $\rightarrow B$, (3) $\rightarrow B \rightarrow C$, (2) $\rightarrow C$ )
(2) (1)(3) (1) $\rightarrow B$, (2) $\rightarrow B \rightarrow C$, (1) $\rightarrow C$, (3) $\rightarrow$ $B \rightarrow C$ )
(2) (3) (1) (1) $\rightarrow B$, (2) $\rightarrow B \rightarrow C$, (3) $\rightarrow B \rightarrow C$, (1) $\rightarrow C$ )
(3)(2)(1) (1) $\rightarrow B$, (2) $\rightarrow B$, (3) $\rightarrow B \rightarrow C$, (2) $\rightarrow$ $C,(1) \rightarrow C$ )

For 4 balls, when the 1st ball is at the bottom, there are 5 ways (first let (1) $\rightarrow B \rightarrow C$, then let (2), (3), (1) correspond to the 5 arrangements of (1), (2), (3) above). When the 2nd ball is at the bottom, there are also 5 ways (first let (1) $\rightarrow B$, (2) $\rightarrow B \rightarrow C$, then let (1), (3), (1) correspond to the 5 arrangements of (1), (2), (3) above, but remove the repeated operation of (1) $\rightarrow B$ in the first step). When the 3rd ball is at the bottom, there are 3 ways: (3) (2)(1)(4), (3)(2)(4)(1), (3)(4)(2)(1). When the 4th ball is at the bottom, there is only 1 way: (4)(3)(2)(1). Therefore, there are a total of 14 ways.