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:
$$3^4-\binom{3}{1}\binom{4}{2}\binom{2}{1}^2+\binom{3}{2}\binom{4}{2}\binom{2}{1}$$

Here is a variation based upon generating functions. Distributing $0$ up to $4$ distinguishable balls to three persons can be represented by
\begin{align*}
  \left(1+t+\frac{1}{2!}t^2+\frac{1}{3!}t^3+\frac{1}{4!}t^4\right)^3\tag{1}
  \end{align*}
The coefficient of $t^4$ multiplied with $4!=24$ in the expanded expression of (1) gives the number of possibilities to distribute four balls to three persons.
Since we do not want that a person gets precisely two balls we use a marker $x$ to indicate this event.
\begin{align*}
  \left(1+t+\color{blue}{x}\cdot \frac{1}{2}t^2+\frac{1}{6}t^3+\frac{1}{24}t^4\right)^3\tag{2}
  \end{align*}
We use the coefficient of operator $[t^j]$ to denote the coefficient of $t^j$ in a series and have to calculate $[t^4x^0]$ in (2).

We obtain
  \begin{align*}
  4!\cdot[t^4x^0]&\left(1+t+x\cdot\frac{1}{2}t^2+\frac{1}{6}t^3+\frac{1}{24}t^4\right)^3\\
  &=4!\cdot[t^4x^0]\left(1+3t+\frac{1}{2}t^2(3x+6)+\frac{1}{6}t^3(18x+9)
  +\frac{1}{24}t^4(18x^2+36x+27)\right)\\
  &=4!\cdot[x^0]\left(\frac{1}{24}(18x^2+36x+27)\right)\\
  &=27
  \end{align*}
We observe out of $18+36+27=81$ possibilities to distribute four distinghishable balls to three persons there are
$\qquad\color{blue}{27} $ possibilities so that none of them has two balls
$\qquad36 $ possibilities  so that exactly one of them has two balls and
$\qquad18 $ possibilities  so that two of them have two balls.

Comment:


*

*In the second line we expand only up to powers of $t^4$ since higher powers do not contribute anything to $[t^4]$. We select the coefficient of $t^4$.

*In the following line we select the coefficient of $x^0$.
