Find the number of ways in which $7$ different balls can be distributed into 4 identical boxes, so that no box remains empty? How to find the number of ways in which $7$ different balls can be distributed into $4$ identical boxes so that no box remains empty?
In this question, I started by finding the number of ways of selecting any $4$ balls and putting them in the identical boxes in one way, and then the remaining balls could be placed in $3^4$ ways.
Can you please help me solving this?
 A: First assume that the boxes are distinct. Then the number of ways of distributing 7 distinct balls into these boxes so that none is empty is, by Principle of Inclusion Exclusion,
$$4^7 - \binom{4}{1}3^7 + \binom{4}{2}2^7 - \binom{4}{3} 1^7 = 8400$$
Now the naming of the boxes can be done in $4! = 24$ ways, the required number is $\dfrac{8400}{24} = 350$.
In fact if we want to distribute $n$ distinct objects into $k$ identical boxes so that no box is empty, the number of ways is given by the Stirling number $S(n,k)$ and 
$$S(n,k) = \frac{1}{k!}\sum_{i=0}^{k-1} \binom{k}{i}(k-i)^n$$
A: You can go $4111, 3211\text {or }2221$, in terms of partitions of $7$...  The boxes are identical so the order doesn't matter. ..
$4111$:  ${7 \choose 4}=35 $
$3211$:  ${7 \choose 3}{4 \choose 2}=35×6=210$
$2221$:  ${7\choose 2}{5\choose 2}{3\choose 2}=21×10×3=630$
Divide $630$ by $3! $ to get 105...
Adding up  $105+210+35=350$
A: Brilliant can be a helpful resource:

A Stirling number of the second kind, denoted as $S(n,r)$ or $\left\{n \atop r\right\}$ is the number of ways a set of $n$ elements can be partitioned into $r$ non-empty sets. In this case, you can use the power of the recurrence relation, which allows one to calculate any value of $S(n,r)$ from other values of $S(n,r)$.

So, here is a list of $S(n,r)$ for $n \leq 7$:
\begin{array}{c|lcr}
\small{n} \ ^{\large{r}} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7\\ \hline
0 & 1 &  &  &  &  & &  &\\
1 & 0 & 1 &  &  &  &  &  \\
2 & 0 & 1 & 1 &  &  &  &  \\
3 & 0 & 1 & 3&  1& &  & &\\
4 & 0 & 1 & 7 & 6& 1 &  &\\
5 & 0 & 1 & 15 & 25& 10 & 1&  & \\
6 & 0 & 1 & 31 & 90& 65 & 15 & 1& & \\
7 & 0 & 1 & 63 & 301 & \color{red}{350} & 140 & 21& 1 & \\
\end{array}
Note that $S(7,4)=\boxed{350}$.
