Consider $3$ digit integers from $000$ to $999$ (i.e., including leading zeroes). Let $n_1$, $n_2$ and $n_3$ refer to the $1$st, $2$nd and $3$rd digit of each of these integers. There are $1000$ such integers. Using the inclusion-exclusion principle, it's simple to determine how many digits have, say, a $1$ in the first position or a $4$ in the second position or a $7$ in the 3rd position - i.e. [$1nn$ OR $n4n$ OR $nn7$]. (There are $271$ such integers.)
However, is there a technique for finding the number of integers that satisfy conditions such as: [$1nn$ OR $n4n$ OR $nn7$] AND [$2nn$ OR $n3n$ OR $nn7$] AND [$1nn$ OR $n7n$ OR $nn9$]
EDIT: sorry, misstated the question originally. By my reckoning, 29 integers meet these conditions. For example, $977$ would meet the conditions despite not having a 1 or 2 in the first position. $977$ meets each of the conditions separately.
This is a trivial example, I'm looking for a pointer to a general technique.
Thanks CL