 Solution

Put $M=2m$ and $a_i=\left(\genfrac{}{}{0ex}{}{d_i}{2}\right)$. Use the configuration model: give vertex $i$ exactly $d_i$ labelled half-edges and choose a uniform perfect matching of all $M$ half-edges. There arepairings, with $(2m{)}_m$ denoting a falling factorial. Each allowed simple graph has exactly ${\prod }_i{d}_i!$ representations: at every vertex, assign its incident graph edges bijectively to its labelled half-edges. Therefore its enumeration reduces to the probability that a pairing is simple and avoids $G_0$.

Count three kinds of prescribed bad patterns. A loop pattern consists of a pair of half-edges at one vertex. A double-edge pattern consists of two half-edges at each of two distinct vertices and one of their two cross-pairings. A forbidden-edge pattern consists of one half-edge at each endpoint of an edge of $G_0$. Let their respective counts in the random pairing be $Z_L,Z_D,Z_F$. Simplicity and avoidance are equivalent to $Z=Z_L+Z_D+Z_F=0$, including exclusion of triple edges because those contain double-edge patterns.

A prescribed collection of $t$ disjoint half-edge pairs occurs with probabilityConsequently the bad-pattern expected values areHere $\lambda =({\sum }_i{a}_i)/m$ and $\mu =M^{-1}{\sum }_{ij\in E(G_0)}{d}_i{d}_j$, as in the PDF. Since $1\le d_i\le \Delta$ and $G_0$ has bounded maximum degree, $M=\Theta (n)$ and all three means are uniformly bounded. For the double-edge estimate, ${\sum }_i{a}_i^2=O(n)$, so its contribution after division by $M^2$ is $o(1)$.

We now give the moment-counting step rather than assume independence. For each fixed $h$, expand the factorial moment $\mathbb{E}(Z{)}_h$ as the sum over ordered distinct bad patterns. Collections on disjoint vertex sets have, by the displayed pairing probability, joint probabilities equal to the product probabilities times $1+O(1/n)$. Their total contribution isDeleting collections with a shared vertex changes the independent-product sum by $O(1/n)$, since all ordinary and forbidden vertex degrees are bounded.

For actual joint probabilities of overlapping patterns, incompatible uses of a half-edge give zero. Compatible collections with no shared paired edge lose at least one free vertex choice, without reducing the number of required pairs, so contribute $O(1/n)$. A shared paired edge can occur only between two double-edge patterns or between a double-edge pattern and a forbidden-edge pattern. In the first case their union requires at least three parallel paired edges: there are $O(n^2)$ endpoint choices and probability $O(n^{-3})$. In the second case the endpoint pair must be an edge of $G_0$: there are only $O(n)$ choices and at least two required pairs, giving $O(n^{-1})$. Adding further patterns in a fixed collection does not remove this deficit: a disjoint addition contributes bounded order, while an addition to an existing cluster has only bounded half-edge choices unless it supplies new free vertices, compensated by its new required pairs. Summing the finitely many overlap types for fixed $h$ therefore gives $o(1)$. Thus, with $\Lambda =\lambda /2+{\lambda }^2/4+\mu$,The Bonferroni inequalities bracket $\mathrm{Pr}(Z=0)$ by the odd and even partial sums of ${\sum }_h(-1{)}^h\mathbb{E}(Z{)}_h/h!$. The means $\Lambda$ are bounded, so the corresponding exponential-series remainders vanish uniformly as the truncation increases. Taking $n\to \infty$ and then the truncation to infinity proves the bounded-degree pairing avoidance estimate $\mathrm{Pr}(Z=0)=e^{-\Lambda }+o(1)$. This does not require $\lambda$ or $\mu$ to converge. The exponential is bounded away from zero, so the additive error is also a relative $o(1)$ error.

Multiplying by the pairing count and dividing by the representation factor givesThe factor $(2m{)}_m$ multiplies the exponential; it is not part of its exponent. The TeX aid misgroups this factor, whereas the original PDF is unambiguous. The bounded-degree proof also covers the stated additional assumption $2m-n\to \infty$.