Let $a$, $b$, $c$ be fixed positive integers. There are $a+b+c$ ducks sitting in a
circle, one behind the other. Each duck picks either rock, paper, or scissors, with $a$ ducks
picking rock, $b$ ducks picking paper, and $c$ ducks picking scissors.
A move consists of an operation of one of the following three forms:

[list]
     [*] If a duck picking rock sits behind a duck picking scissors, they switch places.
     [*] If a duck picking paper sits behind a duck picking rock, they switch places.
     [*] If a duck picking scissors sits behind a duck picking paper, they switch places.
[/list]
Determine, in terms of $a$, $b$, and $c$, the maximum number of moves which could take
place, over all possible initial configurations.
1. **Understanding the Problem:**
   We have \(a\) ducks picking rock, \(b\) ducks picking paper, and \(c\) ducks picking scissors, arranged in a circle. The ducks can switch places based on the rules of rock-paper-scissors. We need to determine the maximum number of moves that can occur.

2. **Initial Configuration:**
   Consider the initial configuration where the ducks are arranged in the order:
   \[
   \underbrace{\text{Rock Rock...Rock}}_{a} \underbrace{\text{Scissors Scissors...Scissors}}_{c} \underbrace{\text{Paper Paper...Paper}}_{b}
   \]
   This configuration is chosen because it maximizes the number of possible moves.

3. **Counting Moves:**
   - Each rock can potentially switch with each scissors.
   - Each paper can potentially switch with each rock.
   - Each scissors can potentially switch with each paper.

4. **Calculating the Maximum Moves:**
   - The number of rock-scissors pairs is \(a \times c\).
   - The number of paper-rock pairs is \(b \times a\).
   - The number of scissors-paper pairs is \(c \times b\).

   Therefore, the total number of possible moves is:
   \[
   a \times c + b \times a + c \times b
   \]

5. **Optimality:**
   To prove this is optimal, consider the number of clockwise pairs of \(\text{Rock}-\text{Scissors}-\text{Paper}\). Initially, there are at most \(a \times b \times c\) of these. However, in each swap of \(x, y\) (where \(x\) is the number of the first type, \(y\) is the number of the second type), it decreases this total by \(z\) (number of the third type). Thus, it decreases by at least \(\min(a, b, c)\). Therefore, there are at most:
   \[
   \frac{a \times b \times c}{\min(a, b, c)} = \max(a \times b, b \times c, c \times a)
   \]
   and we're done.

The final answer is \(\boxed{\max(a \times b, b \times c, c \times a)}\)