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.

• If there is any part that you did not understand in my solution ,you can ask it Jul 8 '21 at 12:13
• Yes I know. Thanks very much for the solution. I'm pretty sure I understand it, I will ask you if I have any parts I don't understand. :) Jul 8 '21 at 13:39

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.
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 , $$6 \times 6\times 6 \times 6\times 6 \times 6\times 6\times6 = 6^8=1679616$$