1. Determine all triples of positive integers $(a, b, n)$ for which

$$
a !+b !=2^{n}
$$

Notation: $k !=1 \times 2 \times \cdots \times k$, so for example $1 !=1$ and $4 !=1 \times 2 \times 3 \times 4=24$.
1. Since $a$ and $b$ play the same role in the equation $a! + b! = 2^n$, for convenience, we assume that $a \leqslant b$. The solutions with $a > b$ can then be found by swapping $a$ and $b$. We examine the possibilities for $a$.

Case $a \geqslant 3$: Since $3 \leqslant a \leqslant b$, both $a!$ and $b!$ are divisible by 3, so $a! + b!$ is also divisible by 3. Since $2^n$ is not divisible by 3 for any value of $n$, there are no solutions.

Case $a = 1$: For $b$, it must hold that $b! = 2^n - 1$. Since $2^n$ is even (because $n \geqslant 1$), $b!$ must be odd. Since $b!$ is divisible by 2 for all $b \geqslant 2$, only $b = 1$ remains. We find $1! = 2^n - 1$, so $n = 1$. The only solution is thus $(a, b, n) = (1, 1, 1)$.

Case $a = 2$: For $b \geqslant 4$, there are no solutions. Indeed, since $b!$ is then divisible by 4, $2^n = b! + 2$ cannot be a multiple of 4, so $2^n = 2$ is the only possibility. But this contradicts the fact that $2^n = b! + 2 \geqslant 24 + 2$.

For $b = 2$, we find $2^n = 2 + 2 = 4$. Thus, $n = 2$ and $(a, b, n) = (2, 2, 2)$ is the only solution.

For $b = 3$, we find $2^n = 2 + 6 = 8$. Thus, $n = 3$ and $(a, b, n) = (2, 3, 3)$ is the only solution. By swapping $a$ and $b$, we also find the solution $(a, b, n) = (3, 2, 3)$.

In total, there are four solutions: $(1, 1, 1)$, $(2, 2, 2)$, $(2, 3, 3)$, and $(3, 2, 3)$.