8. Arrange the six numbers $2, 0, 1, 9, 20, 19$ in any order to form an eight-digit number (the first digit cannot be 0). The number of different eight-digit numbers that can be formed is $\qquad$ .
8. 498 .

Let the set of all permutations of $2, 0, 1, 9, 20, 19$ where the first digit is not 0 be denoted as $A$.
Let $|X|$ denote the number of elements in a finite set $X$.
It is easy to see that $|A|=5 \times 5!=600$.
Let the set of all permutations in $A$ where 2 is followed by 0 and 1 is followed by 9 be denoted as $B$; the set of all permutations in $A$ where 2 is followed by 0 but 1 is not followed by 9 be denoted as $C$; and the set of all permutations in $A$ where 1 is followed by 9 but 2 is not followed by 0 be denoted as $D$.
It is easy to see that $|B|=4!, |B|+|C|=5!$,
$$
|B|+|D|=4 \times 4!.
$$

Thus, $|B|=24, |C|=96, |D|=72$.
Each eight-digit number generated by the permutations in $B$ corresponds to $2 \times 2=4$ permutations in $B$ (in such permutations, 20 can be swapped with “2, 0” and 19 can be swapped with “1, 9”). Similarly, each eight-digit number generated by the permutations in $C$ or $D$ corresponds to 2 permutations in $C$ or $D$.
Therefore, the number of eight-digit numbers that satisfy the conditions is
$$
\begin{array}{l}
|A \backslash(B \cup C \cup D)|+\frac{|B|}{4}+\frac{|C|+|D|}{2} \\
=|A|-\frac{3|B|}{4}-\frac{|C|}{2}-\frac{|D|}{2} \\
=600-18-48-36=498 .
\end{array}
$$