Distributing 4 distinct balls between 3 people 
In how many ways can you distribute 4 distinct balls between 3 people such that none of them gets exactly 2 balls?

This is what I did (by the inclusion–exclusion principle) and I'm not sure, would appreciate your feedback:
$$3^4-\binom{3}{1}\binom{4}{2}\binom{2}{1}^2+\binom{3}{2}\binom{4}{2}\binom{2}{1}$$

One person gets $4$ balls: $3$ ways
One person gets $3$ balls, one person gets $1$: $\dbinom41\times3!=24$
$27$ in total.
