Counting ways to arrange $5$ different balls into $3$ different boxes so that no box remains empty. I get $150$; official answer is $720$.

If the order within the boxes matters, then the solution can be obtained quite easily:

Imagine that the balls are lined up like that: $\circ \: \circ \: \circ \: \circ \: \circ$

• number of arrangements of $\color{blue}{}5$ balls: $\color{blue}{5!}$
• putting in boxes with at least one ball per box corresponds to putting $\color{blue}{2}$ separators into the $\color{blue}{4}$ gaps between the lined up balls: $\color{blue}{\binom{4}{2}}$

All together:

$$\color{blue}{5!}\cdot \color{blue}{\binom{4}{2}} = \boxed{720}$$

Note: This solution assumes that the order within the box matters. You almost got it. There are $2$ distinct "combinations" of balls, namely $(1,1,3)$ and $(1,2,2),$ but you counted the number of possibilities incorrectly. The balls are considered different and so are the boxes, so ordering matters. For the first one, there are ${5\choose 1}\cdot {4\choose 1}\cdot 3!\cdot \frac{3!}{2!}$ arrangements. For the second one, there are ${5\choose 1}\cdot {4\choose 2}\cdot 2!\cdot 2!\cdot \frac{3!}{2!}$ possibilities. How did I get that answer? Well, the ordering of the balls in the boxes must be accounted for. The total matches the desired answer.

The balls are different and the boxes are different we need to consider for the boxes the configurations $(3,1,1),(1,3,1),(1,1,3),(2,2,1),(1,2,2),(2,1,2)$ and therefore

$$3\cdot \binom{5}{3}\cdot 2!+3\cdot \binom{5}{2}\binom{3}{2}=150$$