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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.