8. How many natural numbers not exceeding 1000 are there such that each subsequent digit is greater than the previous one?
8. Among two-digit numbers, there are 36 (see problem 7). Let's count how many such numbers there are among three-digit numbers. From 123 to $129-7$ numbers, from 134 to $139 \cdots 6$ and so on; finally, one number 189. Thus, among three-digit numbers, where the hundreds digit is 1, there are $7+6+\ldots+2+1=28$ such numbers. Among three-digit numbers starting with the digit 2, there are $6+5+\ldots+2+1=21$ such numbers; starting with the digit 3, there are $5+4+3+2+1=15$ such numbers; starting with the digit 4, there are $4+3+2+1=10$ such numbers; starting with the digit 5, there are $3+2+1=6$ such numbers; starting with the digit 6, there are $2+1=3$ such numbers; starting with the digit 7, there is only one number 789. Thus, in total, there are $28+21+15+10+6+$ $+3+1=84$ such three-digit numbers, and together with the two-digit numbers, there will be $36+84=120$.