<s>[INST] Distributing $5$ different balls to $4$ different persons We have to find the number of ways of distributing $5$ different balls to $4$ different persons.
Clearly, the answer is $4^5$ as each ball can be given to any of the $4$ persons. However, I wanted to calculate it using a different method.
I assumed that $a$ balls are given to first person, $b$ to second , $c$ to third and $d$ to fourth person. So we have that $a+b+c+d=5$ where $0 \leq a,b,c,d \leq 5$.
But counting the solutions to the above equation assumes that balls are identical. So I tried to find the number of distributions each permutation of $(a,b,c,d)$ produces. That will be $\displaystyle \binom{5}{a}\cdot \binom{5-a}{b}\cdot \binom{5-a-b}{c} $ which equals $\dfrac{5!}{a!b!c!d!}$.
So, now we need to sum this value over all $a,b,c,d$ satisfying $a+b+c+d=5$. Now there are $\displaystyle \binom{8}{3}=56$ solutions to the equation. So there will be $56$ terms in that summation. So how do we do that?
 [/INST] Consider the partitions $5$ into $4$ parts
\begin{eqnarray*}
(5,0,0,0),(4,1,0,0),(3,2,0,0),(3,1,1,0),(2,2,1,0),(2,1,1,1).
\end{eqnarray*}
These have symmetry factors $4,12,12,12,12,4$ respectively (which adds upto $56$ as you state)
Now the balls can be distributed in each case and multiply in the symmetry factors ...
\begin{eqnarray*}
 4 \times \frac{5!}{5!0!0!0!} + 12 \times \dfrac{5!}{4!1!0!0!} + 12 \times \dfrac{5!}{3!2!0!0!} + 12 \times \dfrac{5!}{3!1!1!0!} + 12 \times \dfrac{5!}{2!2!1!0!} + 4 \times \dfrac{5!}{2!1!1!1!} =1024= 4^5.
\end{eqnarray*}
</s>