Finding number of numbers <= N, containing atleast one of the digits 2,4,6,8
This is much easier than inclusion&ndash;exclusion.  Compute the number of natural numbers smaller than $N$ that contain none of the digits 2, 4, 6, 8 and subtract that from $N$ (or $N+1$ if you're including zero).  To compute the number of naturals that contain none of 2, 4, 6, 8, observe that, if $M\leq N$ (padded with zeroes so $M$ also has $d$ digits) then $M$ and $N$ must agree on their first $i$ digits (for some $0\leq i\leq d$), then the $(i+1)$st digit of $M$ must be less than the $(i+1)$st digit of $N$ and the $(i+2)$nd, ..., $d$th digits of $M$ can be any of 0, 1, 3, 5, 7, 9.