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Similar to an existing thread (Number of ways to choose 4 groups of 4 people from a set of 16 people) I wondered how many combinations into 4 groups could be formed when the groups are ‘non-distinct’ i.e. If group A, B,C and D are swapped, but the persons in these groups stay the same, then it does not count as a new combination.

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You can start with the answer considering the groups to be individually labeled, which as in the cited question is $$\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}$$ but that would count each arrangement $4! = 24$ times since you consider for this problem that applying any permutation of groups $A, B, C, D$ would yield an equivalent grouping. So the answer becomes $$\frac{\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{24}$$

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Label and order your sixteen people in a convenient way, for example by age.

Pick who the three other people who will be in the youngest person's group are. Remove these four people from the pool of available people.

Pick who the three other people who will be in the youngest remaining person's group are. Remove these four people from the pool of available people.

Continue in this fashion until all people are grouped.

You find as a result that there are $\binom{15}{3}\binom{11}{3}\binom{7}{3}$ total groups possible.

(Note: compare to the answer of $\binom{16}{4}\binom{12}{4}\binom{8}{4} / 4!$ whose explanation would rely on a "division by symmetry" argument)

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If groups were distinct, we would have $\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}$.

Now, we have $4!$ ways to permute the groups for a given arrangement of people. Intuitively, it means we count every arrangement $4!$ times. Therefore, we have $\frac{\binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{4!}$ ways.

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