Q: How many positive integers are there whose digits strictly increase from left to right? ($0$ is not allowed to be the leading digit)
Here're my thoughts on the problem.
For one digit numbers.
There are only $9$ such integers that satisfy the problem.
For two digit numbers.
If we start with $1$ as the left most digit, then we have $8$ choices for the one's place. So we have $8$ such numbers.
If we start with $2$ as the left most digit, then we have $7$ choices for the one's place.
If we start with $3$ as the left most digit, then we have $6$ choices for the ones place.
Continuing this process we have $8+7+6+ ... +2+1 = 36$ such numbers.
But for three - digit numbers, I'm stuck. What am I suppose to do, after I've chosen the first digit? I'll have to make sub cases for the second digit, and it'll get more complicated as I go up to 9 - digit numbers.
P.S. I know we have a question such as [Number of strictly increasing and decreasing 4 digit numbers?], but I don't quite get the hints ...
Lets say the number has $k$ digits, all distinct. We notice, that out of all possible permutations of the digits, only one satisfies the condition of the problem, that is, strictly ascending.
If we would try to generate a solution ourselves, we would pick digits from 1 to 9 $k$ times, except we would not be allowed to pick the same digit twice. The number of ways to do this for a fixed $k$ is the number of combinations
$C(n, k) = \frac{n!}{k!(n-k)!}$
where $n=9$ is the total number of digits we choose from
The total number of solutions is sum over number of solutions for a fixed digit length
$S = \sum_{k=1}^{n} C(n, k)$
As mentioned here
https://en.wikipedia.org/wiki/Binomial_coefficient#Sum_of_coefficients_row
There is a theorem for summing up binomial coefficients $C(n,k)$, which states that
$S = 2^n - 1 = 511$ in your case