Four-digit numbers with mandatory digits

Total 4-digit numbers using digits {1, 2, 3} = $3^4 = 81$.

Let $A$ be the set of 4-digit numbers formed without the digit 2 (using only {1, 3}). $|A| = 2^4 = 16$

Let $B$ be the set of 4-digit numbers formed without the digit 3 (using only {1, 2}). $|B| = 2^4 = 16$

$|A \cap B|$ is the set of numbers formed using neither 2 nor 3 (using only {1}). $|A \cap B| = 1^4 = 1$

By Principle of Inclusion-Exclusion, the number of 4-digit numbers lacking 2 or 3 (or both) is: $|A \cup B| = 16 + 16 - 1 = 31$

The number of four-digit numbers where both 2 and 3 appear at least once is: $81 - 31 = 50$.