Avoiding two letters

Direct. Words using only $c$: $1^4=1$ word, $cccc$.

IE check. $|A\cup B|$ where $A$=at least one $a$, $B$=at least one $b$. $|A|=81-2^4=65$, $|B|=65$, $|A\cap B|=81-1=80$ (all words with at least one $a$ or $b$ = total minus $cccc$). $|A\cup B|=65+65-80=50$. Avoid both: $81-50=31$? That gives $31\neq1$.

Hmm, that's wrong. Let me redo: $|A\cap B|$ = words with both $a$ and $b$ (not just at least one of them). $|A\cap B|$ via IE: $|A|+|B|-|A\cup B|$ is circular.

Direct: total words with both $a$ and $b$: $\binom{4}{?}$... easier: total $81$, minus words with no $a$ (16), minus no $b$ (16), plus no $a$ and no $b$ (1): $81-16-16+1=50$. So $|A\cap B|=50-?$... wait, that IS $|A\cup B|$. So $|A\cup B|=50$. Then words with no $a$ and no $b$: $81-50=31$. But that should be words using only $c$: $1$ word. Contradiction!

Ah, my IE is off: words with at least one $a$ OR at least one $b$ = 81 - (words with no $a$ AND no $b$) = 81 - 1 = 80. So $|A\cup B|=80$. Then avoid both $a$ and $b$ = words with no $a$ AND no $b$ = 1.

Conclusion. $1$ word (cccc).

Lesson. IE must be applied carefully to the right sets.