TITLE: Counting ways to arrange $5$ different balls into $3$ different boxes so that no box remains empty. I get $150$; official answer is $720$.
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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
$$
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,s} \right)} {\left[ \matrix{
  s \cr 
  k \cr}  \right]\left\{ \matrix{
  k \cr 
  m \cr}  \right\}\,m^{\,\underline {\,m\,} } }  = m!\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,s} \right)} {\left[ \matrix{
  s \cr 
  k \cr}  \right]\left\{ \matrix{
  k \cr 
  m \cr}  \right\}}  = s!\left( \matrix{
  s - 1 \cr 
  s - m \cr}  \right) = 720\quad \left| \matrix{
  \;s = 5 \hfill \cr 
  \;m = 3 \hfill \cr}  \right.
$$