Let $M$ be a perfect matching on an even number of vertices $n$, and let $\mathbb{S}_n$ be the symmetric group on $n$ vertices. Next, we construct a graph $G_k := ([n], E)$ using $k$ rendomply permuted perfect matchings, i.e., we take univormely at random $k$ permutation from $\mathbb{S}_n$, $\{\pi_i\}_{i=1}^k$, and apply them to the perfect matching $M$ getting the edge set [ E_k := \bigcup_{i=1}^k \pi_i(M). ]

Now, we would like to compute the expectation on the number of cycles in $G_k$, or its first homological group over $\mathbb{Z_2}$. For k=1, it is obviously zero. For $k \geq 3$, we know that the graph is connected with high probability, and since we know that the expectation on the number of edges is $$ \mathbb{E}[|E_k|] = 1 - \left(1 - \frac{1}{n-1}\right)^k \sim \frac{k}{n-1}. $$ We can use Euler characteristic and compute asymptotically that $\text{rank}~H_1(G_k;\mathbb{Z_2}) = O(n)$.

The border case of $k=2$ can be reduced to the Nudel problem geting us $\text{rank}~ H_1(G_2;\mathbb{Z_2}) = O(\log n)$.

My question is, can we compute the same number for dimension 2?

Suppose we have a 2-dimensional complex with a full 1-dimensional skeleton, and suppose we use a Sreiner Triple System (STS) $S \subseteq \binom{[n]}{3}$ on $n$ elements instead of a perfect matching, where $n$ is of the form $6m+1$ or $6m + 3$ for some $m$. Note that in an STS each pair of elements is contained in exactly one triplet.

Chose uniformely at random 3 permutations $\pi_1, \pi_2, \pi_3 \in \mathbb{S}_n$, and construct the follwoing complex $$ K_3 = \binom{[n]}{1}\cup\binom{[n]}{2}\cup\bigcup_{i=1}^{3}\pi_{i}(S). $$

Question: is three a way to estimate the number of ``spheres'' in $K_3$, or more formally $\mathbb{E}[\text{rank}~H_2(K_3;\mathbb{Z_2})]$?