In the paper [B. Derrida, K. Mallick, J. Phys. A 30 (1997) 1031–1046] the following approximation (equation 27 page 6) is done for products of binomial coefficients invoking the Stirling approximation:
$$\frac{{{M}\choose{N+l}} {{M}\choose{N-l}}}{{M\choose N}^2}\approx e^{-Ml^2/N(M-N)}$$
The conditions/assumptions are $M\rightarrow\infty$, $N\leq M$ and $M\geq l\geq1$.
When it comes to binomial coefficients different approximations based on Stirling's formula can be used:
$${n\choose k}\sim\sqrt{\frac{n}{2\pi k (n-k)}}\frac{n^n}{k^k(n-k)^{n-k}}\tag{1}$$
$${n\choose k}\sim\frac{2^n}{\sqrt{n\pi/2}}e^{-(n-2k)^2/(2n)}\tag{2}$$
$${n\choose k}\sim\frac{1}{\sqrt{2\pi k}}\left(\frac{ne}{k}\right)^ke^{-k^2/2n}\tag{3}$$
I have tried using eq. $(2)$ and eq. $(3)$ to recover the result mentioned in [B. Derrida, K. Mallick, J. Phys. A 30 (1997) 1031–1046]. However, I am unable to obtain the same result. Is there an extra assumption to consider, or a different identity I am not aware of when it comes to binomial coefficients?
Using Mathematica the result I get are the following:
from eq. $(2)$ approx $$\frac{{{M}\choose{N+l}} {{M}\choose{N-l}}}{{M\choose N}^2}\approx e^{-4l^2/M}$$
from eq. $(3)$ approx $$\frac{{{M}\choose{N+l}} {{M}\choose{N-l}}}{{M\choose N}^2}\approx e^{-l^2/M}\frac{N(M/N)^{-2N}(M/N+l)^{N+l}(M/N-l)^{N-l}}{\sqrt{(N+l)(N-l)}}$$
