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Section 4 in https://web.stanford.edu/class/archive/cs/cs109/cs109.1206/lectureNotes/LN02_combinatorics.pdf has an explanation of multinomial coefficient to address Selecting multiple groups of objects when objects are all distinct.

I understand the multinomial coefficient having those denominators to reduce overcounting within each group, and that the distinct objects are indistinct within each group from the perspective of group labels assigned to them.

However I don't see any discussion of what happens when not all objects are distinct. For example, I have $A, B, C_1,C_2$ to split into 2 groups of 2 each. If both C are in the same group, the "within group count reduction" works well. However if C1 and C2 are in different groups, like ${A, C_1}$ and ${B, C_2}$, the multinomial coefficient denominator wouldn't be able to recognize that it's the same as ${A,C_2}$ and ${B,C_1}$ and account for that.

Question: Is there any established framework/concept to address this?

Does it also work for cases where there are more than 2 indistinct objects, and possibly spread across more than 2 groups, and spread unevenly?

Han Qi
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  • Assuming that I am interpreting your question correctly, in effect, you are selecting two items from the multiset $~{A,B,C,C},~$ sampling without replacement, where order of selection is deemed irrelevant. The standard method that I employ in such a situation, to enumerate the number of distinct pairs possible, is to break the situation into mutually exclusive cases, depending on whether the pair contains more than $~1~$ C. If both elements of the pair are $~C,~$ there is only one way that can occur. ...see next comment – user2661923 Mar 21 '23 at 10:40
  • Alternatively, if there is less than $~2~$ C's in the pair, then the number of pairs is $~\displaystyle \binom{3}{2},~$ since you are (in effect) choosing two items from a subset of the multiset, where the subset equals $~{A,B,C}.$ – user2661923 Mar 21 '23 at 10:41
  • For more complicated problems, such as how many distinct subsets of size $~10~$ can be formed from the multiset given by ${~A,A,A,A,A,A,B,B,B,B,C,D,E,F,G,H,I,J,K,L},~$ unfortunately, I would feel compelled to consider each of the $~(7 \times 5) = 35~$ mutually exclusive cases separately. That is, the number of $~A$'s $~$ selected is some element in $~{0,1,2,3,4,5,6},~$ and the number of $~B$'s $~$ selected is some element in $~{0,1,2,3,4},~$ – user2661923 Mar 21 '23 at 10:46
  • I emphasize that my intuition in this area is limited, and it is entirely plausible, with the $~10-$ subset problem, that I am overlooking a more elegant approach. See, for example this answer of mine, which (although valid) is very inelegant. The other answer to the exact same question, which is an answer that I was unable to wrap my brain around, let alone originate, provides a very elegant alternative. – user2661923 Mar 21 '23 at 10:50
  • See also https://math.stackexchange.com/questions/2888637/finding-the-amount-partitions-with-gives-sizes-of-a-multiset and https://math.stackexchange.com/questions/315959/finding-all-possible-n-times-n-matrices-with-non-negative-entries-and-given-ro/ – Mike Earnest Mar 21 '23 at 21:48

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