The full permutation of 6 numbers is $5 \cdot 5!=600$. When two 20s appear, the other three numbers are $1,9,19$.
(1) When three empty positions are consecutive, each case corresponds to only 5 permutations;
(2) When two empty positions are consecutive, each pair of cases corresponds to permutations counted once (as shown in the figure below). Therefore, the permutations in this case total $3 \times 5+3 \times 11+6=54$.
Similarly, when two 19s appear, the corresponding permutations also total 54.
When both two 19s and two 20s appear, the corresponding permutations total 6.
Thus, the total number of permutations is $600-2 \times 54+6=498$.
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 $\qquad$