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 $k$ randomly permuted perfect matchings, i.e., we take uniformly at random  $k$ permutations 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 [noodle problem][1] getting 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?

More precisely, we work with a 2-dimensional complex with a full 1-dimensional skeleton. Suppose we use a Steiner 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 uniformly at random 3 permutations $\pi_1, \pi_2, \pi_3 \in \mathbb{S}_n$, and construct the following complex
$$
K_3 = \binom{[n]}{1}\cup\binom{[n]}{2}\cup\bigcup_{i=1}^{3}\pi_{i}(S). 
$$

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


  [1]: https://math.stackexchange.com/questions/1417274/connecting-noodles-probability-question