Number of ways to distribute 5 distinguishable balls between 3 kids such that each of them gets at least one ball 
How many ways are there to distribute 5 distinguishable balls between 3 kids such that each of them gets at least one ball?

My approach is $ \binom{5}{3} 3! $ + $ \binom{2}{2} \binom{3}{2}2!$ which is equal to $66$?
 A: There is a reasonably nice Inclusion/Exclusion way of counting. Forget temporarily about the at least one ball restriction. For each ball, there are $3$ choices for who will get it, for a total of $3^5$.
Now let's count the bad distributions, where one or more kid is left out. We can choose who gets left out in $\binom{3}{1}$ ways. For each choice, there are $2^5$ ways to distribute the balls, for a total of $\binom{3}{1}2^5$. 
However, we have double counted the distributions in which a kid gets all the balls, so we must subtract $3$.
It follows that there are $93$ bad distributions. Subtract this from the $3^5=243$.
Remark: The idea generalizes. 
A: $$\sum_{r_1+r_2+r_3=5,r_i\geq1}\binom{5}{r_1}\binom{5-r_1}{r_2}\binom{5-r_1-r_2}{r_3}=$$
$$=\sum_{r_1+r_2+r_3=5,r_i\geq1}\frac{5!}{r_1!r_2!r_3!}=\frac{5!}{1!1!3!}+\frac{5!}{1!3!1!}+\frac{5!}{3!1!1!}+\frac{5!}{1!2!2!}+\frac{5!}{2!1!2!}+\frac{5!}{2!2!1!}=$$
$$=3(20+30)=150$$
A: This is similar as total Number of onto functions  from $m$ elements to $n$ elements:
$$\sum_{k=0}^n(-1)^k\binom{n}k(n-k)^m\;$$
Given that $m=5$ and $n=3$. 
So, total number of ways such that each of them gets at least one ball
$=3^5-^3C_1(3-1)^5+^3C_2(3-2)^5-^3C_3(3-3)^5$
$=150$
A: It’s not the quickest or most elegant approach, but this problem is straightforwardly done by considering cases. If each kid gets at least one ball, the balls must be distributed either $3$-$1$-$1$ or $2$-$2$-$1$.


*

*$3$-$1$-$1$: There are $\binom31=3$ ways to choose which kid gets $3$ balls, and $\binom53=10$ ways to choose $3$ balls for that kid. There are then $2$ ways to distribute the remaining $2$ balls to the other $2$ kids. This case therefore accounts for $3\cdot10\cdot2=60$ possible distributions.

*$2$-$2$-$1$: There are $3$ ways to choose which kid gets only $1$ ball, and $5$ ways to pick the ball for that kid. There are then $\binom42=6$ ways to choose which $2$ balls go to the next kid in line, and the remaining kid gets the remaining $2$ balls. This case accounts for another $3\cdot5\cdot6=90$ possible distributions.
The correct total, therefore, is $60+90=150$.
Added: It appears to me that you reasoned something like this: First we choose $3$ of the $5$ balls and distribute one of them to each kid; that can be done in $\binom533!$ ways. Then we take both of the remaining $2$ balls, pick $2$ of the $3$ kids, and distribute the last $2$ balls in one of the $2$ possible ways to the $2$ lucky kids. There are several problems with this approach.


*

*You’ve considered only the second of my two cases.

*You’re combining the results of successive choices, not the counts of disjoint cases, so you should be multiplying, not adding: $\binom533!\cdot\binom22\binom322!=60\cdot6=360$.

*You’re overcounting. Suppose that the balls are labelled A, B, C, D, and E. Then you’ve counted the distribution AD | BE | C four times:


*

*once as the result of distributing A, B, and C in the first step and then distributing D and E to the first and second kids in the second step;  

*once as the result of distributing D, B, and C in the first step and then distributing A and E to the first and second kids in the second step;  

*once as the result of distributing A, E, and C in the first step and then distributing D and B to the first and second kids in the second step; and  

*once as the result of distributing D, E, and C in the first step and then distributing A and B to the first and second kids in the second step.


A: Given the variety of methods applied in the previous answers, I can't resist demonstrating one more: an exponential generating function.  Readers not familiar with generating functions can find many resources in the answers to this question: How can I learn about generating functions?
More generally, we might ask the question of how many ways we can distribute $r$ distinguishable balls among three kids, with each kid getting at least one ball.  Let's say the answer is $a_r$, and define the exponential generating function of $a_r$ as
$$f(x) = \sum_{r=0}^{\infty} \frac{1}{r!} a_r x^r$$
The exponential generating function of the number of ways to distribute $r$ balls to a single kid is
$$x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \dots = e^x -1 $$
and the number of ways to distribute $r$ balls to three kids is the "star product" of three such sequences, so
$$f(x) = (e^x-1)^3 = e^{3x} - 3 e^{2x} + 3 e^x -1$$
The answer to the general problem, $a_r$, is $r![x^r]f(x)$, i.e. $r!$ times the coefficient of $x^r$ when $f(x)$ is expanded as an infinite series.  Recalling the infinite series for $e^x$, we see that
$$a_r = 3^r - 3 \cdot 2^r + 3$$
for $r >0$.  In the particular case $r=5$, this yields $a_5 = 150$.
