Putting $8$ distinguishable pairs of $2$ indistinguishable balls in $3$ bins where order doesn't matter 
There are $8$ pairs of $2$ balls, for a total of $16$ balls each. Two balls in the same pair are indistinguishable, but distinguishable from other pairs. How many ways are there to put these balls into $3$ bins where order doesn't matter?

I know that the "Stars and Bars" strategy is to put $n$ indistinguishable objects into $k$ distinguishable bins, but here, some balls are distinguishable and some balls are indistinguishable. I haven't made much progress on the problem. I know how to solve this problem if the balls were all distinguishable and evenly fit into the three bins. However, this problem seems to be a lot trickier. May I have some help? Thanks in advance.
 A: As far as i have understood the boxes are distinguishable and we want to disperse balls , lets call them $(a,a), (b,b),(c,c),(d,d),(e,e),(f,f),(g,g),(h,h)$ without restriction.
In this question ,dispersing different ball pairs are independent from each other , so disperse them separately and multiply them such that
Dispersing $(a,a)$ to three distingusihable bins : $$C(3+2-1,2)=6$$ ways
Dispersing $(b,b)$ to three distingusihable bins: $$C(2+3-1,2)=6$$ ways
This process goes to up to $(h,h)$
Then , $$6 \times 6\times 6 \times 6\times 6 \times 6\times 6\times6 = 6^8=1679616$$
A: Edit: see DavidK's comment, I'm incorrect
It sounds like the balls only become indistinguishable once they are put into a pair. Before pairing them up, we have $16$ distinguishable balls to pair up into $8$ pairs. We can do by choosing $2$ at a time, for a total of $8$ choices, so we calculate ${16\choose 2}{14\choose 2}...{2\choose 2}=\frac{16!}{2^8}$, but the order in which we make these choices doesn't matter, so to account for that we need to divide by $8!$
Now we've put the $16$ balls into $8$ distinguishable pairs, and we can simply count the number of ways to distribute $8$ distinguishable objects into $3$ distinguishable bins, which is $3^8$. So the final count is $3^8 \cdot \frac{16!}{2^88!}$
If the bins are not distinguishable, the problem becomes more complicated and requires the use of Stirling numbers.
