Number of ways to put balls into bins How many ways can I put $12$ balls into $11$ bins, such that no bin contains more than $2$ balls? 
The problem I am having is assigning balls to bins.
 A: I am going to assume that the balls are distinguishable but the bins are not.
Every ball can be a loner (i.e. to be in a bin without any other ball) or a partner (to be in a bin together with another ball). The couple of partners can be from $1$ to $6$, so we have six sub-cases. Assuming that we have $k\in[1,6]$ couples of partners, we have $12-2k$ loners: we may choose them in $\binom{12}{12-2k}=\binom{12}{2k}$ ways. After such choice, we have to build the couples of partners: we may do that in $\frac{(2k)!}{k!2^k}$ ways, so the answer is given by
$$ \sum_{k=1}^{6}\binom{12}{2k}\frac{(2k)!}{k! 2^k} = \color{red}{140151}.$$
A: Here is my answer, going with the assumption that having identical bins means that having two in the first and one in each other is the same as having two in the last and one in each other. 
Essentially, we are figuring out different ways to add up to $12$ using eleven numbers, were the numbers being added are various combinations of $0$'s, $1$'s, and $2$'s. 
We have:
$$2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 $$
$$2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 $$
$$2 + 2+ 2 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 0 $$
$$2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 0 + 0 + 0 $$
$$2 + 2 + 2 + 2 + 2 + 1 + 1 + 0 + 0 + 0 + 0 $$
$$2 + 2 + 2 + 2 + 2 + 2 + 0 + 0 + 0 + 0 + 0 $$
So the total number of ways to arrange the balls in bins is $6$ ways, which was stated in the comments by Christian Blatter. 
If the balls/bins are distinguishable, I would suggest looking at the answer supplied by Jack D'Aurizio
