Let $G(n,p)$ be a random graph on $n$ vertices, where $p$ is the probability of an edge between any two vertices, and $pn \to c>0$ as $n\to \infty$.
Let $H=C_3 \sqcup C_3$ be two non-intersecting triangles.
Prove that $X_n$, the number of subgraphs of $G$ isomorphic to $H$, satisfies $X_n(H) \xrightarrow{d} \frac{1}{2}Z(Z-1)$, where $Z \sim \text{Pois}(c^3/6)$.
We can use the Poisson limit theorem for graphs, which states that if $A$ is a strictly balanced graph and $np \to c > 0$ as $n\to \infty$, then $X_n(A) \xrightarrow{d} \text{Pois}\left(\frac{c^{v_A}}{\text{aut}(A)}\right)$, where $\text{aut}(A)$ is the number of automorphisms
So, for each triangle: $X_n(C_3)\sim \text{Pois}\left(\frac{c^3}{6}\right)$.
However, I don't know how to extend this result to $C_3 \sqcup C_3$.
We have $$\binom{X_n(C_3)}{2} = X_n(C_3 \sqcup C_3) + X_n(B) + X_n(K_4^-)$$ where $B$ is the butterfly graph and $K_4^-$ is the diamond graph. The reasoning is simple: if you choose two distinct copies of $C_3$ in a graph, together they form either $C_3 \sqcup C_3$, $B$, or $K_4^-$.
In $G(n,p)$ where $p \sim \frac cn$ as $n \to \infty$, it is quick to check that $\mathbb E[X_n(B)] \to 0$ and $\mathbb E[X_n(K_4^-)] \to 0$ as $n \to \infty$. Therefore $X_n(B) \xrightarrow d 0$ and $X_n(K_4^-) \xrightarrow d 0$ as $n \to \infty$, letting us solve for the limiting distribution of $X_n(C_3 \sqcup C_3)$.