If $12$ distinct balls are distributed to $8$ numbered cells, what is the probability that there is no empty cell? I am trying to solve this question:
Let us assume we are distributing 12 different balls between 8 numbered cells, so that each distribution of balls into cells is obtained with equal probability. What is the probability that there is no empty cell?
At first I gave to each ball an unique ID, such that:
1 = ball number 1,
2 = ball number 2, and so on.
Now i defined $\Omega =\left\{ \left( x_{1},\ldots ,x_{12}\right) | \forall i,x_{i}\in \left[ 8\right] \right\} =\left[ 8\right] ^{12}$, and so the event we want to compute is:
$A=\{ \left( x_{1},\ldots ,x_{12}\right) \in \Omega | \forall i\in \left[ 8\right] \exists j\in \left[ 12\right] ,x_{j}=i \}$.
At first I tried to work with $A^{c}$, but then I noticed that I have duplicates, so I tried to compute $|A|$ directly.
I said, in order for no cell to be empty, we will choose 8 balls out of the 12 and distribute them into the eight cells, so that each cell contains exactly one ball. Then, we will distribute the remaining four balls into the eight cells and finish.
So let's do the math:

*

*Choose 8 unique balls out of 12 is $\begin{pmatrix} 12 \\ 8 \end{pmatrix}$

*Distribute the balls to the cells is $8!$

*Then, distribute the remaining four is $8^{4}$
So, overall we have: $|A| = \begin{pmatrix} 12 \\ 8 \end{pmatrix} \cdot 8! \cdot 8^{4}$
Now, $|\Omega| = 8^{12}$, so finally we get:
$\mathbb{P}(A) = \dfrac{\begin{pmatrix} 12 \\ 8 \end{pmatrix}\cdot 8!\cdot 8^{4}}{8^{12}} = 1.18961 >1$.
I don't understand what I'm doing wrong, so would glad for help.
 A: Note that in your calculations you are counting the next senerio twice, ball 1 was chosen as one of the "filing" balls and was put in cell 1, or ball 1 wasn't chosen as one of the "filing" balls but still was put in cell 1.
To fix this divide by the number of duplicates. While noting that this number changes in correspondence to how mach was added to each cell (4 balls in 4 different cells is different then 4 balls in the same cell).
A: Each ball has $8$ choices for boxes, so total choices $=8^{12}$ which will form the denominator $D$
So we shall just focus on the number of ways, the numerator $N$
and the required probability will be $\frac{N}{D}$

*

*The simplest way is to use Stirling numbers of the second kind, which places distinct objects into identical boxes, and permute the boxes, hence
$8!*S(12,8) = 6411968640$

*If you don't have a calculator for Stirling numbers,the usual way is to apply inclusion-exclusion

*The direct way may appear tedious, but can be made  semi-mechanical using permutations using the formula [Lay down pattern] $\times$ [Permute pattern]
$5 + 1 + 1 + 1 + 1 + 1 + 1 + 1:\;\Large\frac{12!}{5!}\times\frac{8!}{7!}$
$4 + 2 + 1 + 1 + 1 + 1 + 1 + 1:\;\Large\frac{12!}{4!2!}\times\frac{8!}{6!}$
$3 + 3 + 1 + 1 + 1 + 1 + 1 + 1:\;\Large\frac{12!}{3!3!}\times\frac{8!}{2!6!}$
$3 + 2 + 2 + 1 + 1 + 1 + 1 + 1:\;\Large\frac{12!}{3!2!2!}\times\frac{8!}{2!5!}$
$2 + 2 + 2 + 2 + 1 + 1 + 1 + 1:\;\Large\frac{12!}{2!2!2!2!}\times \frac{8!}{4!4!}$
Add up to get $N= 6411968640 $
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
& \color{#44f}{\sum_{c_{1} = \color{red}{\Large1}}^{\infty}{1 \over 12}
\sum_{c_{2} = \color{red}{\Large1}}^{\infty}{1 \over 12}\ldots
\sum_{c_{8} = \color{red}{\Large1}}^{\infty}{1 \over 12}}
\bracks{z^{12}}z^{c_{1}\ +\ c_{2}\ +\ \cdots\ +\ c_{8}}
\\[5mm] = & \
{1 \over 12^{8}}\bracks{z^{12}}\pars{\sum_{c = 1}^{\infty}z^{c}}^{8} =
{1 \over 12^{8}}\bracks{z^{12}}\pars{z \over 1 - z}^{8} =
{1 \over 12^{8}}\bracks{z^{4}}\pars{1 - z}^{-8}
\\[5mm] = & \
{1 \over 12^{8}}{-8 \choose 4}\pars{-1}^{-4} =
{1 \over 12^{8}}{11 \choose 4}\pars{-1}^{4} =
{1 \over 12^{8}}\, 330
\\[5mm] = &\
\bbx{\color{#44f}{55 \over 71663616}} \approx 7.6747 \times 10^{-7}
\end{align}
