Question: How to count unlabeled balls and labeled boxes case

each box can have more than one balls or some box may have no balls

use example , 10 labeled boxes, 8 unlabeled balls

welcome to use polynomial counting

better have closed formula or generating function

would like to know Prob(a labelled box have 3 balls) too

Answer: 
For a $2\times2$ matrix $A$, if it is not diagonalisable, it must be similar to a Jordan block, i.e. $A=P(\lambda I+J)P^{-1}$ where $P$ is invertible, $\lambda$ is the repeated eigenvalue of $A$ and $J=\pmatrix{0&1\\ 0&0}$. Since $J^2=0$,
\begin{align\*}
A^n
&=P(\lambda I+J)^nP^{-1}\\
&=P(\lambda^nI+n\lambda^{n-1}J)P^{-1}\\
&=P\left[n\lambda^{n-1}(\lambda I+J)-(n-1)\lambda^nI\right]P^{-1}\\
&=n\lambda^{n-1}A-(n-1)\lambda^nI\tag{1}
\end{align\*}
and you may calculate $a\_n$ as
$a\_n=(0,1)\ A^{\color{red}{n-1}} \pmatrix{a\_2\\ a\_1}$.

Note that $(1)$ works for $2\times2$ matrices only (and it hinges on the fact that $J^2=0$). For larger-sized matrices, their Jordan blocks may have different sizes and hence you should calculate the power blockwise, as suggested by Git Gud's answer. Also, as mentioned in the answer of Clement C., you can solve the linear recurrence directly without turning it into a matrix form.
There is no need to be that complicated. 
Here is the trick.
an+2=4an+1−4an, then an+2-2an+1=2(an+1-2an).
So we can conclude that {an+1-2an} is an geometric progression.
 (an+1-2an)/(a2-2a1)=2^(n-1),
an+1-2an=-2^(n-1),
2an-4an-1=-2^(n-2)\*2=-2^(n-1),
4an-1-8an-2=-2^(n-3)\*4=-2^(n-1),
.......
2^(n-1)\*a2-2^n\*a1=-2^(n-1),
sum up ,we get an+1-2^n=-n\*2^(n-1),
so an=2^(n-1)-(n-1)\*2^(n-2)=(3/4-n/4)\*2^n.