smc

Case 1: monotonous numbers with digits in ascending order There are ${\sum }_{n=1}^9\left(\genfrac{}{}{0ex}{}{9}{n}\right)$ ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, $\varnothing$ (the empty set) isn't included because it doesn't generate a number. The sum is equivalent to ${\sum }_{n=0}^9\left(\genfrac{}{}{0ex}{}{9}{n}\right)-\left(\genfrac{}{}{0ex}{}{9}{0}\right)=2^9-1=511.$ Case 2: monotonous numbers with digits in descending order There are ${\sum }_{n=1}^{10}\left(\genfrac{}{}{0ex}{}{10}{n}\right)$ ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros. However, $\varnothing$ (the empty set) still isn't included because it doesn't generate a number. The sum is equivalent to ${\sum }_{n=0}^{10}\left(\genfrac{}{}{0ex}{}{10}{n}\right)-\left(\genfrac{}{}{0ex}{}{10}{0}\right)=2^{10}-1=1023.$ We discard the number 0 since it is not positive. Thus there are $1022$ here. Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are $511+1022-9=\boxed{1524}$ monotonous numbers.