Find all the positive integers $a, b, c$ such that $$a! \cdot b! = a! + b! + c!$$
We need to find all positive integers \(a, b, c\) such that:
\[ a! \cdot b! = a! + b! + c! \]

We will consider different cases based on the value of \(a\).

1. **Case 1: \(a = 1\)**
   \[
   1! \cdot b! = 1! + b! + c! \implies b! = 1 + b! + c!
   \]
   Simplifying, we get:
   \[
   c! + 1 = 0
   \]
   This is a contradiction since \(c!\) is always a positive integer. Therefore, there are no solutions for \(a = 1\).

2. **Case 2: \(a = 2\)**
   \[
   2! \cdot b! = 2! + b! + c! \implies 2b! = 2 + b! + c!
   \]
   Simplifying, we get:
   \[
   b! = 2 + c!
   \]
   Since \(b!\) must be a factorial, we need to check possible values of \(c\).

   - **Subcase 2.1: \(c = 2\)**
     \[
     b! = 2 + 2! = 2 + 2 = 4
     \]
     This implies \(b = 3\) since \(3! = 6\), which is a contradiction. Therefore, \(c = 2\) is not a solution.

   - **Subcase 2.2: \(c = 3\)**
     \[
     b! = 2 + 3! = 2 + 6 = 8
     \]
     This implies \(b = 4\) since \(4! = 24\), which is a contradiction. Therefore, \(c = 3\) is not a solution.

   Thus, there are no solutions for \(a = 2\).

3. **Case 3: \(a \geq 3\)**
   Without loss of generality, assume \(b \geq a\). We can rewrite the equation as:
   \[
   a! \cdot b! = a! + b! + c!
   \]
   Rearranging, we get:
   \[
   c! = a! \cdot b! - a! - b!
   \]
   Since \(a \geq 3\), we have:
   \[
   c! = a! \cdot b! - a! - b! \geq 4b!
   \]
   This implies \(c > b\). Also, since \(c > b\), \(b!\) must divide \(a! \cdot b! - b! - c!\), which implies \(b!\) divides \(a!\). Therefore, \(a \geq b\), which forces \(a = b\).

   Substituting \(a = b\) into the equation, we get:
   \[
   a! \cdot a! = 2a! + c! \implies a!^2 = 2a! + c!
   \]
   Dividing both sides by \(a!\), we get:
   \[
   a! = 2 + \frac{c!}{a!}
   \]
   We need to check possible values of \(c\) relative to \(a\).

   - **Subcase 3.1: \(c = a + 1\)**
     \[
     a! = 2 + \frac{(a+1)!}{a!} = 2 + (a+1)
     \]
     Simplifying, we get:
     \[
     a! = a + 3
     \]
     Checking for \(a = 3\):
     \[
     3! = 6 \quad \text{and} \quad 3 + 3 = 6
     \]
     This is a valid solution. Therefore, \(a = 3, b = 3, c = 4\) is a solution.

   - **Subcase 3.2: \(c = a + 2\)**
     \[
     a! = 2 + \frac{(a+2)!}{a!} = 2 + (a+2)(a+1)
     \]
     Simplifying, we get:
     \[
     a! = a^2 + 3a + 4
     \]
     Checking for \(a = 4\):
     \[
     4! = 24 \quad \text{and} \quad 4^2 + 3 \cdot 4 + 4 = 24
     \]
     This is not a valid solution.

   - **Subcase 3.3: \(c = a + 3\)**
     \[
     a! = 2 + \frac{(a+3)!}{a!} = 2 + (a+3)(a+2)(a+1)
     \]
     Simplifying, we get:
     \[
     a! = a^3 + 6a^2 + 11a + 8
     \]
     Checking for \(a \leq 8\) manually reveals no solutions.

Therefore, the only solution is \(a = 3, b = 3, c = 4\).

The final answer is \( \boxed{ (a, b, c) = (3, 3, 4) } \)