Count the $7$ digit numbers whose digits increase strictly from left
to right.
I have seen some posts here, but I am not sure it's still clear to me how to calculate this kind of problems. We have the set $\{0,1,2,3,4,5,6,7,8,9\}$ and the leading digit cannot be $0$. This is what I've noticed and understand. Can you help me to continue?
Best Answer
The trick of combinatorics is to figure out how to count the complicated thing in terms of things that we already know how to count.
In this case, the order of the digits is already set, so if I gave you seven digits from the set $\{1,2,3,4,5,6,7,8,9\}$, you'd be able to construct exactly one seven digit number with ascending digits from them. And the same thing happens in reverse -- if I gave you a seven digit number with ascending digits, you'd be able to come up with exactly one seven-element subset. Therefore, the answer to your question is exactly the number of seven-element subsets of $\{1,2,3,4,5,6,7,8,9\}$. The good news is that this is simple to express as a binomial combination, it's just $\binom97=36$.