Finding non-bouncy numbers under 10^x
This is not a review but an expanded comment.

Brute force is bad. Do some math first.

Consider a non-decreasing number of length \$n\$ formed by  exactly \$k\$ given digits. It has exactly \$k - 1\$ transitions from a digit to a larger one, and there are exactly \$n - 1\$ places where the transition may happen, so there are \$\binom{n-1}{k-1}\$ such numbers. The \$\binom{10}{k}\$ ways to pick \$k\$ digits give a subtotal of \$\binom{n-1}{k-1}\binom{10}{k}\$, and a grand total is \$\sum_{k=1}^{10}\binom{n-1}{k-1}\binom{10}{k}\$.

So the problem nails down to efficient calculation of binomial coefficients, which in this case is fairly straightforward, since you never need more than 10 factors.