# Asymptotic Stirling approximation of binomial coefficients

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 $$\frac{{{M}\choose{N+l}} {{M}\choose{N-l}}}{{M\choose N}^2}\approx e^{-l^2/M}\frac{N(M/N)^{-2N}(M/N+l)^{N+l}(M/N-l)^{N-l}}{\sqrt{(N+l)(N-l)}}$$

• in the referenced paper the approximation is done in page 6 equation 27. They do use a different notation from the one I used. Commented Feb 27, 2023 at 18:51

$$\frac{N!^2(M-N)!^2}{(N+l)!(N-l)!(M-N+l)!(M-N-l)!}\\ =\frac{N(N-1)\cdots(N-l+1)}{(N+1)(N+2)\cdots(N+l)}\cdot\frac{(M-N)(M-N-1)\cdots(M-N-l+1)}{(M-N+1)(M-N+2)\cdots(M-N+l)}\\ \frac{\left(1-\frac1N\right)\cdots\left(1-\frac{l-1}N\right)}{\left(1+\frac1N\right)\cdots\left(1+\frac lN\right)}\cdot\frac{\left(1-\frac1{M-N}\right)\cdots\left(1-\frac{l-1}{M-N}\right)}{\left(1+\frac1{M-N}\right)\cdots\left(1+\frac l{M-N}\right)}\;.$$
$$\sum_{k=1}^{l-1}k+\sum_{k=1}^lk=\frac{(l-1)l}2+\frac{l(l+1)}2=l^2$$
$$-l^2\left(\frac1N+\frac1{M-N}\right)=-l^2\frac M{N(M-N)}\;.$$