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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.

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  • $\begingroup$ If there is any part that you did not understand in my solution ,you can ask it $\endgroup$
    – Bulbasaur
    Jul 8 '21 at 12:13
  • $\begingroup$ 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. :) $\endgroup$
    – Star Light
    Jul 8 '21 at 13:39
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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 , $$6 \times 6\times 6 \times 6\times 6 \times 6\times 6\times6 = 6^8=1679616$$

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