We choose 5 out of the 85 spots for the c’s to be placed: $ \binom{85}{5}$
Each of the remaining 80 characters have 3 possible letters to be, which are $a$, $b$, or $d$: $ 3 \cdot 3 \cdot \ldots \cdot 3 = 3^{80}$
Multiplying everything up, we get $ \binom{85}{5} \cdot 3^{80} $