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: 
[Jordan Normal Form](http://en.wikipedia.org/wiki/Jordan_normal_form), [this](https://math.stackexchange.com/questions/326688/why-the-nth-power-of-a-jordan-matrix-involves-the-binomial-coefficient) and [this](http://en.wikipedia.org/wiki/Jordan_normal_form#Powers) should make it clear.

Regarding the matrix in your problem you get [this](http://www.wolframalpha.com/input/?i=jordan+normal+form+%7B%7B4%2C-4%7D%2C%7B1%2C0%7D%7D) decomposition.

The above links yield $J^{n-1}=\begin{pmatrix}2^{n-1} & {n-1\choose 1}2^{n-2}\\ 0 & 2^{n-1}\end{pmatrix}$. 

Now you can use $M^k=(SJS^{-1})^k=SJ^kS^{-1}$.
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.