The problem isn’t overlaps: it’s restricting the values to the range from $0$ through $9$. You’re looking for the number of solutions in non-negative integers to the equation
with the added requirement that $x_k\le 9$ for $k=1,\dots,6$: each of those solutions defines one of the numbers that you’re trying to count, and vice versa. The number of solutions without the upper bound restriction is, as you say,
However, some of those solutions require ‘digits’ greater than $9$. Start by counting the solutions in non-negative integers to $(1)$ that have $x_1\ge 10$. Those are clearly in one-to-one correspondence with solutions in non-negative integers to
where $y_1=x_1-10$, and you know how to compute their number. Call that number $n_1$. There are just as many solutions to $(1)$ that violate the upper bound on $x_2$, just as many again that violate the upper bound on $x_3$, and so on, so a better approximation to the solution to the original problem is
Of course this overcorrects, since it’s possible for a solution to $(1)$ to violate two of the upper bound constraints at once. You’ll have to calculate the number $n_2$ of solutions that have $x_1\ge 10$ and $x_2\ge 10$ and then add back the appropriate multiple of $n_2$. And since it’s actually possible to violate as many as three of the constraints at once, you’ll have to make one further inclusion-exclusion correction.