For how many ordered triplets of three positive integers is it true that their product is four more than twice their sum? 
1. Let the three positive integers be \(a, b, c\) such that \(a \leq b \leq c\). We are given the equation:
   \[
   abc = 2(a + b + c) + 4
   \]

2. We can rewrite the equation as:
   \[
   abc = 2(a + b + c) + 4
   \]

3. Since \(a, b, c\) are positive integers, we can start by considering the smallest possible values for \(a\) and \(b\) and then solve for \(c\).

4. Let's start with \(a = 1\):
   \[
   1 \cdot b \cdot c = 2(1 + b + c) + 4
   \]
   Simplifying, we get:
   \[
   bc = 2 + 2b + 2c + 4
   \]
   \[
   bc = 2b + 2c + 6
   \]
   Rearranging, we get:
   \[
   bc - 2b - 2c = 6
   \]
   Adding 4 to both sides, we get:
   \[
   bc - 2b - 2c + 4 = 10
   \]
   Factoring, we get:
   \[
   (b-2)(c-2) = 10
   \]

5. The pairs \((b-2, c-2)\) that satisfy this equation are:
   \[
   (1, 10), (2, 5), (5, 2), (10, 1)
   \]
   This gives us the pairs \((b, c)\) as:
   \[
   (3, 12), (4, 7), (7, 4), (12, 3)
   \]

6. Since \(a \leq b \leq c\), we need to check which of these pairs satisfy \(1 \leq b \leq c\):
   \[
   (3, 12), (4, 7)
   \]

7. Now, let's consider \(a = 2\):
   \[
   2 \cdot b \cdot c = 2(2 + b + c) + 4
   \]
   Simplifying, we get:
   \[
   2bc = 4 + 2b + 2c + 4
   \]
   \[
   2bc = 2b + 2c + 8
   \]
   Dividing by 2, we get:
   \[
   bc = b + c + 4
   \]
   Rearranging, we get:
   \[
   bc - b - c = 4
   \]
   Adding 1 to both sides, we get:
   \[
   (b-1)(c-1) = 5
   \]

8. The pairs \((b-1, c-1)\) that satisfy this equation are:
   \[
   (1, 5), (5, 1)
   \]
   This gives us the pairs \((b, c)\) as:
   \[
   (2, 6), (6, 2)
   \]

9. Since \(a \leq b \leq c\), we need to check which of these pairs satisfy \(2 \leq b \leq c\):
   \[
   (2, 6)
   \]

10. Finally, let's consider \(a = 3\):
    \[
    3 \cdot b \cdot c = 2(3 + b + c) + 4
    \]
    Simplifying, we get:
    \[
    3bc = 6 + 2b + 2c + 4
    \]
    \[
    3bc = 2b + 2c + 10
    \]
    Rearranging, we get:
    \[
    3bc - 2b - 2c = 10
    \]
    This equation does not factor nicely into integers, so there are no solutions for \(a = 3\).

11. Therefore, the valid ordered triplets \((a, b, c)\) are:
    \[
    (1, 3, 12), (1, 4, 7), (2, 2, 6)
    \]

The final answer is \(\boxed{3}\).