【Analysis】Let $\overline{A B C D E}+\overline{\mathrm{abcde}}=99999$
First, determine if there is a carry,
$$
\mathrm{A}+\mathrm{B}+\mathrm{C}+\mathrm{D}+\mathrm{E}+\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}
$$
$=0+1+2+3+4+5+6+7+8+9=45=9+9+9+9+9$, no carry. (It can also be determined that there is no carry by using $8+9=17$, which cannot be 19)
Next, form five pairs of 9: $9=0+9=1+8=2+7=3+6=4+5$ $\mathrm{a}+\mathrm{b}$ and $\mathrm{b}+\mathrm{a}$ are considered the same, so let 0 be in $\mathrm{ABCDE}$, and the different permutations of $\mathrm{ABCDE}$ are $2 \times 2 \times 2 \times 2 \times 4 \times 24=1536$ (ways).
When $\overline{A B C D E}$ is fixed, it uniquely determines $\overline{\mathrm{abcde}}$.
There are a total of 1536 ways to fill in.
19. Fill in the following squares with $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ respectively, so that the sum of the two five-digit numbers is 99999. How many different addition equations are there? (Consider $a+b$ and $b+a$ as the same equation)
$\square \square \square \square \square+$ $\square$ $=99999$