Solution: Let's divide the numbers into groups:

1) Numbers in which all digits are different - there are a total of $9 \cdot 9 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=9 \cdot 9$ !.
2) Numbers in which two non-zero digits are the same and do not stand in the highest place. There are 9 such digits, pairs of positions where these digits can stand - $\frac{9 \cdot 8}{2}=36$, the remaining digits can be arranged in the remaining 8 positions in $8 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=8 \cdot 8$ ! ways. In total, there are such numbers $36 \cdot 8 \cdot 9!$.
3) Numbers in which two zero digits are the same. There are pairs of positions where these digits can stand $-\frac{9 \cdot 8}{2}=36$, the remaining digits can be arranged in the remaining 8 positions in $9 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=9$ ! ways. In total, there are such numbers $36 \cdot 9$ !.
4) Numbers in which two non-zero digits are the same, and one of them stands in the highest place. There are 9 such digits, pairs of positions where these digits can stand - 9, the remaining digits can be arranged in the remaining 8 positions in $9 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=9$ ! ways. In total, there are such numbers $81 \cdot 9$ !.

In total, there are $(9+36 \cdot 8+36+81) \cdot 9!=414 \cdot 9!$ special numbers.

Answer: $414 \cdot 9!$
3. Let's call a natural number special if one of its digits can be replaced by another digit so that all digits in the resulting number are distinct. Numbers in which all digits are already distinct are also considered special. How many special ten-digit numbers exist? (20 points)