, 2022

Neither $a$ nor $b$ can be $6$ or $20$ because either of those numbers would make an arithmetic progression with either the least three numbers or the greatest three numbers. Therefore $a$ and $b$ must be chosen from the remaining $22$ values between $7$ and $29$ not including $20$. Setting $(a, b) = (7, 9)$ results in the only arithmetic progression that contains two of the three least values. Because $20$ is already excluded, no arithmetic progression can be formed using exactly two of the three greatest numbers. All other possible arithmetic progressions must start with one of $3$, $4$, or $5$ and end with one of $30$, $40$, or $50$. The difference of the starting and ending terms must be divisible by $3$, which yields three potential pairs for starting and ending numbers of the arithmetic progression of length four. Using $3$ and $30$, only $(a, b) = (12, 21)$ gives an arithmetic progression. Using $4$ and $40$, only $(a, b) = (16, 28)$ gives an arithmetic progression. The arithmetic sequence starting with $5$ and ending with $50$ would contain $20$ and has already been excluded. Therefore the requested number of ordered pairs is $\binom{22}{2} - 3 = 228$.