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The parameter relations give $\mu =\log (n/c)$ andFor each fixed integer $r\ge 1$, take an $r$-vertex set $S$ and let $X_S$ count triangles meeting $S$. The event that every vertex of $S$ is counted by $T$ is exactly $X_S=0$.

The number of such triangles isThus their total expected count is ${\mu }_S=N_S{p}^3=r\mu +O_r(n{p}^3)=r\mu +o(1)$.

Distinct triangles have dependent containment events only when they share an edge. If that common edge meets $S$, there are $O_r(n)$ choices for it and $O(n^2)$ choices for the two other vertices. If the common edge avoids $S$, both other vertices must lie in $S$, giving only $O_r(n^2)$ pairs. Each pair needs five edges. Consequently its ordered Janson dependency sum isJanson inequality gives the upper bound $\mathrm{Pr}(X_S=0)\le \exp (-{\mu }_S+{\Delta }_S/2)$. For the lower bound, the avoidance of each triangle is a decreasing event. Repeated application of Harris' inequality givesThe error $n^2{p}^6=O((\log n{)}^2/n^2)$ also tends to zero. The bounds match multiplicatively:Summing over ordered distinct vertices yields the factorial moments

Here is an explicit justification that these moments determine the limit. For every integer $t\ge 0$, the finite alternating sumsare upper bounds for $\mathrm{Pr}(T=t)$ when $L$ is even and lower bounds when $L$ is odd. This follows by applying the Bonferroni inequalities to $\left(\genfrac{}{}{0ex}{}{T}{t}\right){\sum }_j(-1{)}^j(\genfrac{}{}{0ex}{}{T-t}{j})$; the full sum is precisely the indicator of $T=t$. First let $n\to \infty$ for a fixed truncation and use the factorial-moment limits. Then let the even and odd truncations grow. Both bounds tend toHence every point probability has the Poisson distribution limit, proving the Poisson threshold for vertices lying in no triangle:This is also the factorial-moment criterion for Poisson convergence, here justified rather than left as an unproved implication.

Take $c>0$ and $e>0$, as required for the displayed sparse scaling and the later normalization. A strictly balanced graph satisfies $e(F)/v(F)<e/v$ for every proper nonempty subgraph $F\subset H$. This forces $H$ to be connected and to have no isolated vertices: otherwise a component of at least the average density, or deletion of the isolated vertices, violates strict balance. Copies here are subgraphs, not induced subgraphs.

For fixed $r$, the factorial moment ${\mathbb{E}}_r(X_H)$ counts ordered $r$-tuples of distinct copies all present in the binomial random graph. The contribution of vertex-disjoint copies is exactlyThe factor $a^{-r}$ removes the multiple embeddings of each copy coming from its graph automorphisms. For an overlapping tuple, let $U$ be its union. There are $O(n^{v(U)})$ placements of each fixed union type, and the probability its edges are present is $p^{e(U)}$. When adding a copy meeting the earlier union in $f$ vertices and $g$ edges, the exponent $v(U)-(v/e)e(U)$ changes byIt is strictly negative if this intersection is a proper nonempty subgraph, and zero only for an empty intersection or a whole copy. In particular the first vertex-overlapping copy has a proper nonempty intersection: the preceding copies are vertex-disjoint, and a connected copy contained in their union would have to equal one of them, contrary to distinctness. Its contribution is therefore strictly negative; later additions cannot undo that loss. Only finitely many union types occur for fixed $r$, so their total contribution is $o(1)$. We have proved all the factorial moment limits, and the allowed factorial-moment criterion for Poisson convergence givesThis is the Poisson limit for strictly balanced subgraph counts.

The pendant-edge graph has density $7/5$, whereas its proper $K_4$ core has density $6/4$. Thus $K_4^{+}$ is not strictly balanced. The complete graph $K_4$ is strictly balanced, with 24 graph automorphisms, so at $p=c{n}^{-2/3}$ its copy count $Z$ tends to $\mathrm{Po}(c^6/24)$.

Every $K_4^{+}$ copy has a unique four-vertex clique core and one selected edge joining it to a fifth vertex. For any four-set $C$, let $D_C$ count the ambient edges between $C$ and its complement. Then $D_C\sim \mathrm{Bin}(4(n-4),p)$, with mean ${\mu }_n\sim 4c{n}^{1/3}$. It is independent of the internal edges determining whether $C$ is a clique. Choose ${\delta }_n=n^{-1/12}$. A Chernoff bound and a union bound giveTherefore, simultaneously for every actual core, its number of pendant extensions is $(1+o(1))4c{n}^{1/3}$. Extra ambient edges cause no overcount: each selected seven-edge subgraph has exactly one core and one pendant edge. Summing over the cores yieldsbecause $Z$ is uniformly tight. This is the pendant extensions of sparse clique copies mechanism. For $0<\epsilon <1$, neither endpoint of $(k-\epsilon ,k+\epsilon )$ is an integer, and its only nonnegative integer is $k$. Convergence in distribution consequently givesThe positive-$c$ assumption matters: the quotient printed in the question is undefined when $c=0$.

Let $Y$ be the number of isolated vertices and set $\lambda =e^{-c}$. For fixed $k\ge 1$, the falling factorial $(Y{)}_k$ counts ordered $k$-tuples of distinct isolated vertices. A specified tuple is isolated exactly when all edges incident to its vertices are absent. There are $k(n-k)+\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ such edges, soSince $np=\log n+c$ and $n{p}^2=o(1)$, taking logarithms givesThus every fixed factorial moment tends to ${\lambda }^k$. We use the standard factorial-moment criterion for Poisson convergence: nonnegative integer-valued random variables whose factorial moments converge, for every fixed order, to ${\lambda }^k$ converge in distribution to $\mathrm{Poisson}(\lambda )$. Consequently $Y$ has that limiting distribution, and its mass at each fixed integer converges. In particular,This establishes the limiting probability of at least one isolated vertex, rather than merely a bound based on its expectation.

Here $v,e,a$ are the numbers of vertices, edges and graph automorphisms of $H$. Vertex-disjoint ordered $r$-tuples contribute $(n{)}_{rv}{p}^{re}/a^r\to (c^e/a{)}^r$ to the factorial moment. Adding a copy overlapping the previous union in $f$ vertices and $g$ edges changes the power of $n$ by $-f+(v/e)g\le 0$, strictly when the intersection is a proper nonempty subgraph. The first overlap is strict because $H$ is connected and the previous copies are disjoint. Finitely many overlap types therefore contribute $o(1)$. The factorial-moment criterion for Poisson convergence gives the result.

For a fixed set of $r$ vertices, exclude all triangles meeting that set. Their total expectation is $r\log (n/c)+o(1)$. Distinct triangle pairs sharing an edge contribute $O_r(n^3{p}^5)=o(1)$ to the Janson dependency sum. Harris' inequality and Janson inequality bound the avoidance probability between matching exponentials, giving $(c/n{)}^r(1+o(1))$. The factorial-moment criterion for Poisson convergence then gives the displayed law.