Counting ways to arrange $5$ different balls into $3$ different boxes so that no box remains empty. I get $150$; official answer is $720$. 
In how many ways can $5$ different balls be arranged into $3$ different boxes so that no box remains empty?

My Approach: Acc to me, since there are $2$ ways to arrange the balls ie,
$(1,1,3)$ and $(1,2,2)$ the total number of methods would be 
$$ \frac{C^{5}_1\cdot C^4_1\cdot  C^3_3 \cdot 3!}{2!} +\frac{C^{5}_1\cdot C^4_2\cdot  C^2_2 \cdot 3!}{2!} = 150$$
By arranging the balls and then multiply by $3!$ as amount of ways $3$ different boxes can be arranged and dividing by $2!$ because one value is same.
Correct Answer: However, the correct answer is given as $$\frac{5! \cdot 3!}{2!}+ \frac{5! \cdot 3!}{2!}$$

What is wrong with my approach and why does the answer vary?

Edit: After reading all the answers, I also think that the question is not stated clearly and the order within the box must be important. It was taken from Skills In Mathematics algebra by Dr. SK Goyal (Chapter Permutation and Combination) if someone wants to look further upon the question.
 A: 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:
$$\color{blue}{5!}\cdot \color{blue}{\binom{4}{2}} = \boxed{720}$$
A: 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.
A: 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
$$3\cdot \binom{5}{3}\cdot 2!+3\cdot \binom{5}{2}\binom{3}{2}=150$$
A: I think that your approach is correct (also see my comment on your question).
Another approach:
Number the boxes with $1,2,3$ and for $i=1,2,3$ let $A_i$ denote the set of possibilities such that box $i$ remains empty.
Without any constraints there are $3^5$ possibilities.
Then to be found is: $$|A_1^{\complement}\cap A_2^{\complement}\cap A_2^{\complement}|=3^5-|A_1\cup A_2\cup A_3|$$
Applying inclusion/exclusion and symmetry we find that this equals:$$3^5-3|A_1|+3|A_1\cap A_2|=3^5-3\cdot2^5+3=150$$
A: Let's put it general.
How many ways there are to "put" $s$ different balls into $m$ different boxes ( for the moment assuming no empty/full limitation) ?
Well, the answer is that it depends on how we realize the "putting" process, or more rigorously 
on how we define the space of equi-probable results.
And omitting this definition we can get quite different results.
We can take the balls sequentially and launch them into the boxes and consider equiprobable each of the $m^s$ different landings.
That is the same as considering equiprobable each  of the $m^s$ functions from $\{1,2, \cdots,s\}$ to $\{1,2, \cdots,m\}$.
If  we draw the "occupancy" histograms of the boxes ordered in succession, we will have all the possible histograms differing in quantity and/or identity
of the balls in each box.
But the order of the balls inside each box is fixed: ball $k$ would have landed there before any successive one.
Imagine the boxes as  tennis balls cans.
If instead we arrange the balls in every possible order (permutations of $s$ balls) and then apply the stars and bars arrangement 
(which practically means to "throw" the boxes into the balls), the number of ways to do that will be 
$$
s!\left( \matrix{
  s + m - 1 \cr 
  s \cr}  \right) = m^{\,\overline {\,s\,} } 
$$
where $x^{\,\overline {\,k\,} } $ represent the  Rising Factorial.
In this case the occupancy histograms will be all those differing by quantity and/or identity and/or order of the balls in each can.
So coming to your problem, in the first case we have
$$
m^{\,s}  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,\min (s,m)} \right)} {\left\{ \matrix{
  s \cr 
  k \cr}  \right\}m^{\;\underline {\,k\,} } }  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,\min (s,m)} \right)} {k!\left\{ \matrix{
  s \cr 
  k \cr}  \right\}\left( \matrix{
  m \cr 
  k \cr}  \right)} 
$$
where the Stirling Number of 2nd kind ${s\brace k}$
denote in fact the number of ways to partition a set of $s$ objects into $k$ non-empty subsets and
the Falling Factorial $m^{\;\underline {\,k\,} }$
the number of ways to assign the subsets to the boxes.   
With $s=5, \; m=3$ this becomes
$$
\left\{ \matrix{
  s \cr 
  m \cr}  \right\}m^{\;\underline {\,m\,} }  = \left\{ \matrix{
  s \cr 
  m \cr}  \right\}m! = 150
$$
and of course there are many formulations to render the above, including yours.
In the second case, instead, the rising factorial decomposes as
$$
m^{\,\overline {\,s\,} }  = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,s} \right)} {\left[ \matrix{
  s \cr 
  k \cr}  \right]m^{\,k} } \; = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,s} \right)} {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,k} \right)} {\left[ \matrix{
  s \cr 
  k \cr}  \right]\left\{ \matrix{
  k \cr 
  j \cr}  \right\}\,m^{\,\underline {\,j\,} } } } \;
$$
So
$$
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,s} \right)} {\left[ \matrix{
  s \cr 
  k \cr}  \right]\left\{ \matrix{
  k \cr 
  m \cr}  \right\}\,m^{\,\underline {\,m\,} } }  = m!\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,s} \right)} {\left[ \matrix{
  s \cr 
  k \cr}  \right]\left\{ \matrix{
  k \cr 
  m \cr}  \right\}}  = s!\left( \matrix{
  s - 1 \cr 
  s - m \cr}  \right) = 720\quad \left| \matrix{
  \;s = 5 \hfill \cr 
  \;m = 3 \hfill \cr}  \right.
$$
