Decide whether there exist infinitely many triples $(a,b,c)$ of positive integers such that all prime factors of $a!+b!+c!$ are smaller than $2020$.

[i]Pitchayut Saengrungkongka, Thailand[/i]
1. **Assume Without Loss of Generality (WLOG)**: We can assume \( a \geq b \geq c \) because the factorial function is strictly increasing for positive integers. This assumption simplifies our analysis without loss of generality.

2. **Bound on \( c \)**: Since we want all prime factors of \( a! + b! + c! \) to be smaller than 2020, consider the smallest prime greater than 2020, which is 2027. If \( c \) were greater than or equal to 2027, then \( c! \) would be divisible by 2027 (since 2027 is a prime number), and thus \( c! \) would have 2027 as a prime factor. This would contradict our requirement that all prime factors of \( a! + b! + c! \) are smaller than 2020. Therefore, \( c \) must be less than 2027.

3. **Bound on \( b \)**: Given \( c < 2027 \), we now consider \( b \). If \( b \) were greater than or equal to 2027, then \( b! \) would also be divisible by 2027, introducing 2027 as a prime factor in \( a! + b! + c! \). Thus, \( b \) must also be less than 2027.

4. **Bound on \( a \)**: Similarly, if \( a \) were greater than or equal to 2027, then \( a! \) would be divisible by 2027, introducing 2027 as a prime factor in \( a! + b! + c! \). Thus, \( a \) must also be less than 2027.

5. **Conclusion**: Since \( a, b, \) and \( c \) must all be less than 2027, there are only finitely many possible values for \( a, b, \) and \( c \). Specifically, \( a, b, \) and \( c \) can each take any integer value from 1 to 2026. Therefore, there are only finitely many triples \((a, b, c)\) such that all prime factors of \( a! + b! + c! \) are smaller than 2020.

\[
\boxed{\text{No}}
\]