Determine all triples of positive integers $(a, b, n)$ that satisfy the following equation: $a! + b! = 2^n$
1. **Initial Considerations:**
   - We need to find all triples of positive integers \((a, b, n)\) such that \(a! + b! = 2^n\).
   - Note that \(2^n\) is a power of 2, and thus it is even. Therefore, \(a! + b!\) must also be even.

2. **Case Analysis:**
   - If both \(a\) and \(b\) are greater than 3, then \(a! + b!\) is a multiple of 3 (since any factorial of a number greater than 3 includes 3 as a factor). However, \(2^n\) is never a multiple of 3 for \(n \geq 1\). Hence, at most one of \(a\) or \(b\) can be greater than 3.

3. **Assume \(a \geq b\):**
   - This assumption simplifies our analysis since we can consider the cases for \(b\) and then determine \(a\).

4. **Case \(b = 1\):**
   - The equation becomes \(a! + 1! = 2^n\), which simplifies to \(a! + 1 = 2^n\).
   - If \(a \geq 2\), then \(a! + 1\) is odd (since \(a! \geq 2! = 2\) is even), but \(2^n\) is even, leading to a contradiction.
   - Therefore, \(a = 1\), giving the solution \((1, 1, 1)\).

5. **Case \(b = 2\):**
   - The equation becomes \(a! + 2! = 2^n\), which simplifies to \(a! + 2 = 2^n\).
   - Rearranging, we get \(a! = 2^n - 2\).
   - We need to check values of \(a\) such that \(a! = 2^n - 2\) is a factorial and a power of 2 minus 2.
   - For \(a = 2\), \(2! = 2\), so \(2 = 2^n - 2\) implies \(2^n = 4\) and \(n = 2\). This gives the solution \((2, 2, 2)\).
   - For \(a = 3\), \(3! = 6\), so \(6 = 2^n - 2\) implies \(2^n = 8\) and \(n = 3\). This gives the solution \((3, 2, 3)\).

6. **Case \(b = 3\):**
   - The equation becomes \(a! + 3! = 2^n\), which simplifies to \(a! + 6 = 2^n\).
   - Rearranging, we get \(a! = 2^n - 6\).
   - We need to check values of \(a\) such that \(a! = 2^n - 6\) is a factorial and a power of 2 minus 6.
   - For \(a = 3\), \(3! = 6\), so \(6 = 2^n - 6\) implies \(2^n = 12\), which is not a power of 2. Thus, no solution for \(a = 3\).
   - For \(a = 2\), \(2! = 2\), so \(2 = 2^n - 6\) implies \(2^n = 8\) and \(n = 3\). This gives the solution \((2, 3, 3)\).

7. **Conclusion:**
   - The only valid solutions are \((1, 1, 1)\), \((2, 2, 2)\), \((3, 2, 3)\), and \((2, 3, 3)\).

The final answer is \(\boxed{(1, 1, 1), (2, 2, 2), (3, 2, 3), (2, 3, 3)}\).