8. Arrange the 6 numbers $2,0,1,9,20,19$ in any order to form an 8-digit number (the first digit cannot be 0), then the number of different 8-digit numbers that can be generated is $\qquad$ .
Answer: 498.
Solution: Let the set of all permutations of $2,0,1,9,20,19$ where the first digit is not 0 be denoted as $A$.
It is easy to see that $|A|=5 \times 5!=600$ (here and below, $|X|$ represents the number of elements in the finite set $X$).
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 $C$; 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 $|B|=4!, |B|+|C|=5!, |B|+|D|=4 \times 4!$, i.e., $|B|=24, |C|=96, |D|=72$.
Each 8-digit number generated by the permutations in $B$ corresponds exactly 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 8-digit number generated by the permutations in $C$ or $D$ corresponds exactly to 2 permutations in $C$ or $D$. Therefore, the number of 8-digit numbers that meet the conditions is
$$
\begin{array}{c}
|A \backslash(B \cup C \cup D)|+\frac{|B|}{4}+\frac{|C|+|D|}{2} \\
=|A|-\frac{3|B|}{4}-\frac{|C|}{2}-\frac{|D|}{2}=600-18-48-36=498 .
\end{array}
$$