Consider a set of four elements $$ Y^0 = \{ y_1, y_2, y_3, y_4 \} $$ Let $Y^1$ be the set that includes all pairwise combinations of distinct elements of $Y^0$ $$ Y^1 = \{ y^1_1, \dots, y^1_6 \} := \{ \{y_1, y_2\}, \{y_1, y_3\}, \{y_1, y_4\}, \{y_2, y_3\}, \{y_2, y_4\}, \{y_3, y_4\} \} $$ By construction, each element of $Y^1$ contains each element of $Y^0$ at most once. Now let's build $Y^2$ as the pairwise combinations of distinct elements of $Y^1$ \begin{align} Y^2 &= \{ y^2_1 \dots, y^2_{15} \} := \\ & := \{ \ \{\{y_1, y_2\}, \{y_1, y_3\}\}, \ \{\{y_1, y_2\}, \{y_1, y_4\}\}, \ \{\{y_1, y_2\}, \{y_2, y_3\}\}, \dots, \ \{\{y_2, y_4\}, \{y_3, y_4\}\} \} \end{align} 12 elements of $Y^2$ are such that one element of $Y_0$ appears twice. 3 elements of $Y_2$ contain only distinct elements of $Y^0$.

At each step $N$ we build $Y^N$ as the set of pairwise combinations of distinct elements of $Y^{N-1}$.

**Question**: is there a way to know, given $N$, how many elements of $X_N$ include how many repetitions of elements of $X_0$? The information I am looking for is something like "*$\ell$ elements of $X_N$ are such that they include an element of $X_0$ three times, another distinct element of $X_0$ two times and a third distinct element of $X_0$ one time*" and so on, I am not interested in which specific elements of $X_0$ is repeated.

**Addendum:** I am adding a picture that should describe the generation of the nested sets better.

"set of 2-element subsets of $X_0$". And each element of $X_1$contains(not "includes") every element of $X_0$at mostonce. $\endgroup$ – YCor Jun 6 '20 at 17:394more comments