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?

Work backwards:
Suppose you already gave out $a$ balls to the first person and $b$ balls to the second.   You now have $\binom{5-a-b}{c}$ ways of giving $c$ balls to the third person and the remaining balls to the fourth person.  Summing this up for all possible value of $c$ gives: $\binom{5-a-b}{0}+\binom{5-a-b}{1}+...+\binom{5-a-b}{5-a-b}=2^{5-a-b}$.
Suppose you gave out $a$ balls to the first person.  You now have $\binom{5-a}{b}$ ways of giving $b$ balls to the second person, and for each of those you are left with $2^{5-a-b}$ ways of giving out balls to the remaining two people.  Summing this up for all values of $b$ gives:
$\binom{5-a}{0}2^{5-a}+\binom{5-a}{1}2^{4-a}+...+\binom{5-a}{5-1}2^{0}=3^{5-a}$.
Lastly, summing for $a$, you have $\binom{5}{0}3^5+\binom{5}{1}3^4+\binom{5}{2}3^3+\binom{5}{3}3^2+\binom{5}{4}3^1+\binom{5}{5}3^0=4^5$.
This uses the identity:
$$\sum_{k=0}^X\binom{X}{k}Y^k=\sum_{k=0}^X\binom{X}{X-k}Y^k=(Y+1)^X$$
This identity comes from the fact that this sum is what you get when you expand the binomial $(1+Y)^X$
