Show With High Probability, No Vertex Belongs to More than One Triangle I am working on a random graphs problem, which is stated as follows:

Suppose that $p = d/n$, where $d$ is constant. Prove that with high probability (w.h.p.), no vertex belongs to more than one triangle.

For each $i \in \left[\binom{n}{3}\right] \cup \{0\}$, I define $X_{i}$ as the set of vertices in the graph that belong to $i$ triangles. I believe that I want to show that $Pr[v \in X_{0} \cup X_{1}] \to 1$ as $n \to \infty$. 
I begin by fixing a vertex $v$ and deriving $Pr[v \in X_{0}]$ and $Pr[v \in X_{1}]$. If $v$ belongs to no triangles, it may have up to $n-1$ neighbors. If $v$ has at least two neighbors, none of them may be adjacent. This yields the probability:
$$Pr[v \in X_{0}] = (1-p)^{n-1} + (n-1)p(1-p)^{n-2} + \sum_{i=2}^{n-1} \binom{n-1}{i} p^{i} (1-p)^{n-i-1+\binom{i}{2}}$$
Similarly, if $v \in X_{1}$, it necessarily has at least two neighbors, and exactly two of its neighbors are adjacent. This yields the probability:
$$Pr[v \in X_{1}] = \sum_{i=2}^{n-1} \binom{n-1}{i} \binom{i}{2} p^{i+1} (1-p)^{n-i-2+\binom{i}{2}}$$
Am I correct that I want to show $Pr[v \in X_{0} \cup X_{1}] \to 1$ as $n \to \infty$? Would it be possible to get a nudge in the right direction on how to show the desired result (w.h.p., no vertex belongs to more than one triangle)?
I greatly appreciate everyone's time and effort to help me!
 A: There is a simple argument showing that asymptotically no vertex is involved is more than one triangle. 
Let $g$ be a subgraph containing at least two triangles, $k$ its number of vertices, $m$ its number of edges. 
This is subgraph of excess at least 1 (i.e. $m-k \geqslant 1$).
Now what is the probability to find $g$ in a graph where all edges are independently taken with probability $c \cdot n^{-1}$? 
There are ${n \choose k}=\Theta(n^k)$ ways to find such a subset of vertices. Each subset has at least $m$ edges with probability $\Theta(n^{-m})$. 
So overall the probability to find $g$ is $\Theta(n^{k-m})= O(n^{-1})$, so is vanishing.
A: I think I have an answer. I would love some feedback or critiques of ways I could improve this (especially if I'm off base).
Define $X_{i}$ as above. We have:
$$Pr[v \in \bigcup_{i=2}^{\binom{n}{3}} X_{i}] = 1 - Pr[v \in X_{0}] - Pr[v \in X_{1}]$$
It suffices to show:
$$\lim_{n \to \infty} Pr[v \in \bigcup_{i=2}^{\binom{n}{3}} X_{i}] = \lim_{n \to \infty} (1 - Pr[v \in X_{0}] - Pr[v \in X_{1}]) = 0$$
We have: $Pr[v \in X_{0}]$ and $Pr[v \in X_{1}]$ as in the OP. Evaluating $Pr[v_{0} \in X_{1}] + Pr[v \in X_{1}]$:
$$Pr[v_{0} \in X_{1}] + Pr[v \in X_{1}] = (1-p)^{n-1} + (n-1)p(1-p)^{n-2} + \sum_{k=2}^{n-1} \binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}} \biggr[ \biggr( \binom{k}{2} - 1 \biggr)p + 1 \biggr]$$
Evaluating:
$$
\lim_{n \to \infty} Pr[v_{0} \in X_{1}] + Pr[v \in X_{1}] = 
\lim_{n \to \infty} (1-p)^{n-1} + 
\lim_{n \to \infty} (n-1)p(1-p)^{n-2} + 
\lim_{n \to \infty} \sum_{k=2}^{n-1} \binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}} \biggr[ \biggr( \binom{k}{2} - 1 \biggr)p + 1 \biggr]
$$
 We have $\lim_{n \to \infty} (1-p)^{n-1} +  \lim_{n \to \infty} (n-1)p(1-p)^{n-2} = e^{-d}(1 + d)$. 
Now applying the linearity property of the limit, we obtain:
$$\lim_{n \to \infty} \sum_{k=2}^{n-1} \binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}} \biggr[ \biggr( \binom{k}{2} - 1 \biggr)p + 1 \biggr] =  $$
$$\sum_{k=2}^{\infty} \lim_{n \to \infty}\binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}} \biggr[ \biggr( \binom{k}{2} - 1 \biggr)p + 1 \biggr]$$
Distributing $\biggr[ \biggr( \binom{k}{2} - 1 \biggr)p + 1 \biggr]$ and applying the linearity property of the limit again, we obtain:
$\lim_{n \to \infty}\binom{n-1}{k}p^{k+1}(1-p)^{n-k-2+\binom{k}{2}} (\binom{k}{2}-1) = 0$.
So:
$$\sum_{k=2}^{\infty} \lim_{n \to \infty}\binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}} \biggr[ \biggr( \binom{k}{2} - 1 \biggr)p + 1 \biggr] = $$
$$\sum_{k=2}^{\infty} \lim_{n \to \infty}\binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}}$$
And $\lim_{n \to \infty}\binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}} = \dfrac{d^{k}}{e^{d}k!}$. 
Thus:
$$\sum_{k=2}^{\infty} \lim_{n \to \infty}\binom{n-1}{k}p^{k}(1-p)^{n-k-2+\binom{k}{2}} = \sum_{k=2}^{\infty} \dfrac{d^{k}}{e^{d} k!} = \dfrac{e^{d} - d - 1}{e^{d}}$$
Recall $\lim_{n \to \infty} (1-p)^{n-1} +  \lim_{n \to \infty} (n-1)p(1-p)^{n-2} = e^{-d}(1 + d)$. Adding these limits up yields: $1$.
And so: $\lim_{n \to \infty} (1 - Pr[v \in X_{0}] - Pr[v \in X_{1}]) = 0$, proving the desired result.
