<s>[INST] Given $n$ distinct items, distribute them to $m$ distinct bins such that no bin is empty One solution can be solved by using inclusion-exclusion principle:
Let $A$ be the group that all items exist and to compute $A'$ then reduce it from the universe.
Then I thought about another way to solve it but I think it's not accurate -
let $A$ be the group that all bins have at least $1$ item.
The way I computed $A$:
Let's choose $m$ items from the $n$ items which is $\binom{n}{m}$ and distribute one item to each bin such that none of the bins is empty, there are $m!$ ways to do that.
Then I have $n-m$ items that can go to any bin, so it's $m^{n-m}$.
Total - $\binom{n}{m} \cdot m! \cdot m^{n-m}$
Any idea why it's not correct or what's wrong with this way of approaching the problem.
 [/INST] Let's look at the case $n = 5$ and $m = 3$.  We will solve the problem in two ways, then examine why your approach does not work.
Method 1:  We use the Inclusion-Exclusion Principle.
There are three choices for each of the five objects, so the objects can be distributed in $3^5$ ways.  From these, we must subtract those distributions in which one or more bins is left empty.
There are three ways to select a bin that will be left empty and $2^5$ ways to distribute the five objects to the remaining bins.  
However, if we subtract $3 \cdot 2^5$ from the total, we will have subtracted the three cases in which two bins are left empty twice, once for each way we could designate one of the empty bins as the bins to be left empty.  We only want to subtract those cases once, so we must add them back.
Thus, the number of ways of distributing five distinct objects to three distinct bins so that no bin is left empty is
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 150$$
Method 2:  This method is practical in the case $n = 5$, $m = 3$ since the difference between $n$ and $m$ is small.  However, it will be useful in illustrating why your method did not work.
We can express $5$ as a sum of three positive integers in two ways:
\begin{align*}
5 & = 3 + 1 + 1\\
  & = 2 + 2 + 1
\end{align*}
Three items are placed in one bin and one object each is placed in the other bins:  Choose which of the three bins will receive three items.  Choose three of the five objects to place in that bin.  Arrange the remaining two items in the remaining two bins.  There are
$$\binom{3}{1}\binom{5}{3}2! = 60$$
such distributions.
One item is placed in one bin and each of the other bins receives two objects:  Choose which of the three bins receives a single object.  Choose which of the five objects is placed in that bin.  Choose which two of the remaining four objects to be placed in the leftmost open bin.  The remaining two objects must be placed in the remaining open bin.  There are
$$\binom{3}{1}\binom{5}{1}\binom{4}{2} = 90$$
such distributions.
Total: Since these cases are mutually exclusive and exhaustive, there are a total of 
$$\binom{3}{1}\binom{5}{3}2! + \binom{3}{1}\binom{5}{1}\binom{4}{2} = 60 + 90 = 150$$
ways to distribute five distinct objects to three distinct bins so that no bin is left empty.
What is wrong with your approach?
Suppose we wish to distribute the numbers $1, 2, 3, 4, 5$ to bins $A$, $B$, and $C$.  
If we place the numbers $1$, $2$, and $3$ in bin $A$, $4$ in bin $B$, and $5$ in bin $C$, you count that case three times, once for each of the three numbers you could designate as the object that goes in bin $A$.
$$
\begin{array}{c c c c}
\text{bin $A$} & \text{bin $B$} & \text{bin $C$} & \text{additional objects in bin $A$}\\ \hline
1 & 4 & 5 & 2, 3\\
2 & 4 & 5 & 1, 3\\
3 & 4 & 5 & 1, 2
\end{array}
$$
If we place the numbers $1$ and $2$ in bin $A$, $3$ and $4$ in bin $B$, and $5$ in bin $C$, you count that case four times, twice for each way you could designate one of the two numbers in bin $A$ as the object that goes in bin $A$ and twice for each way you could designate one of the two numbers in bin $B$ as the object that goes in bin $B$.
$$
\begin{array}{c c c c}
\text{bin $A$} & \text{bin $B$} & \text{bin $C$} & \text{additional object in bin $A$} & \text{additional object in bin $B$}\\ \hline
1 & 3 & 5 & 2 & 4\\
1 & 4 & 5 & 2 & 3\\
2 & 3 & 5 & 1 & 4\\
2 & 4 & 5 & 1 & 3
\end{array}
$$ 
Notice that your approach yields the answer
$$\color{red}{\binom{5}{3}3! \cdot 3^2 = 10 \cdot 6 \cdot 9 = 540}$$
and that 
$$\color{red}{\binom{3}{1}}\binom{3}{1}\binom{5}{3}2! + \color{red}{\binom{2}{1}\binom{2}{1}}\binom{3}{1}\binom{5}{1}\binom{4}{2} = \color{red}{3} \cdot 60 + \color{red}{4} \cdot 90 = \color{red}{180} + \color{red}{360} = \color{red}{540}$$
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