Probability that at least 2 edges of $\Gamma_{n,N}$ shall have a point in common In the classic paper of Erdos,Renyi On the evolution of random graphs[page 7] ,it is argued that the probability that at least 2 edges of $\Gamma_{n,N}$ shall have a point in common is given by $1-\frac{{n\choose 2N}(2N)!}{2^N N!{{n \choose 2}\choose N}} = \mathcal{O}(\frac{N^2}{n})$.  
Here $n$ is order and $N$ is size of the random graph.  
Can anyone please describe how the asymptotic formula is derived? 
IN general, how to derive such results involving factorials and powers?
 A: An important condition on $N$ in that paper is that $N = o(n^{1/2}).$ This allows us to more easily deal with the binomial coefficient terms. Namely, since $N = o(n^{1/2})$, we have that
$$
{n \choose 2N} = (1+o(1)) \frac{n^{2N}}{(2N)!},
$$
and
$$
{{n \choose 2} \choose N} = (1+o(1)) \frac{ {n \choose 2}^N}{N!}.
$$
But we will need to know more about these $o(1)$ terms and so we have to go deeper. To do this, let's expand the first binomial expression:
\begin{align*}
{n \choose 2N} &= \frac{1}{(2N)!} \prod_{i=0}^{2N-1} (n - i) \\&= \frac{n^{2N}}{(2N)!} \prod_{i=0}^{2N-1} \left( 1 - \frac{i}{n} \right).
\end{align*}
Further, note that
\begin{align*}
\prod_{i=0}^{2N-1} \left( 1 - \frac{i}{n} \right) &= \exp \sum_{i=0}^{2N-1} \log \left(1- \frac{i}{n}\right) \\ & = \exp \sum_{i=0}^{2N-1} \left( - \frac{i}{n} + O\left(\frac{N^2}{n^2}\right) \right) \\&= \exp \left( \frac{-2N(2N-1)}{n} + O\left( \frac{N^3}{n^2}\right)\right) \\&= \exp \left(O \left( \frac{N^2}{n} \right)\right) = 1+ O\left( \frac{N^2}{n} \right).
\end{align*}
Thus 
$$
{n \choose 2N} = \left(1+O\left(\frac{N^2}{n}\right)\right)\frac{n^{2N}}{(2N)!}.
$$
The asymptotic expression for the other binomial expression can be shown using a similar method to get
$$
{{n \choose 2} \choose N} = \left(1+O\left(\frac{N^2}{{n\choose 2}}\right)\right)\frac{{n \choose 2}^{N}}{N!}.
$$
Putting these all together, we have that
\begin{align*}
\frac{ { n \choose 2N} (2N)!} {2^N N! {{n \choose 2} \choose N}} &= \left(1+O\left(\frac{N^2}{n}\right)\right) \frac{ n^{2N}}{2^N {n \choose 2}^N} \\&= \left(1+O\left(\frac{N^2}{n}\right)\right) \frac{n^{2N}}{2^N n^N (n-1)^N/2^N} \\&= \left(1+O\left(\frac{N^2}{n}\right)\right) \frac{1}{\left(1-\frac{1}{n}\right)^N}.
\end{align*}
Now note that $\left(1-\frac{1}{n}\right)^N = 1+O\left(\frac{N}{n}\right).$
As a consequence,
$$
\frac{ { n \choose 2N} (2N)!} {2^N N! {{n \choose 2} \choose N}} = 1+O\left(\frac{N^2}{n}\right).
$$
In this case, we did not need to approximate the factorials since each factorial was divided out. If you want an asymptotic formula for a factorial, you would use Stirling's approximation. Namely,
$$
n! = (1+o(1)) \sqrt{2\pi \, n} \, \left(\frac{n}{e}\right)^n.
$$
If you wanted to approximate further, that $o(1)$ error term above is actually on the order of $1/n$. 
