Can we solve for (or approximate) $m$ from $ \frac{(2n-m)!}{(n-m)!} = c$? I need to solve the following equation for $m$:
$$
\frac{(2n-m)!}{(n-m)!} = c
$$
where $m$ and $n$ are natural numbers in the order of 10 with $n \ge m$ and $c$ is a real-valued constant (edit: it's a natural number actually, HT @Kaind). Any chance this can be solved/approximated analytically?
 A: I believe there is no analytical solution for $(m,n) \in\Bbb N^2$. For numerical solutions, the time complexity of the problem is $\mathcal{O}(n^2)$ (the problem is a two dimensional problem for $(m,n)$ ).
This method here provides a solution with the time complexity around $\mathcal{O}(n)$.
First, we extend the problem from $(m,n) \in\Bbb N^2$ to $(m,n) \in \Bbb R^2$ and solve the new problem
$$\frac{\Gamma(2n-m+1)}{\Gamma(n-m+1)}=c \tag{1}$$
with the function Gamma defined here  (for $n\in \Bbb N$, we have $\Gamma(n) = (n-1)!$)
For the sake of simplicity, let's denote $(x,y) = (n,n-m+1)$, then
\begin{align}
(1)&\iff \Gamma(x+y)=c\Gamma(y) \\
&\iff x =\Gamma^{-1}(c\Gamma(y))-y\tag{2}
\end{align}
According to this result, the inverse Gamma function $\Gamma^{-1}(z)$ can be approximated by
$$\Gamma^{-1}(z) \approx \frac{\ln \left(\frac{z+d}{\sqrt{2\pi}}\right)}{W \left( \frac{\ln \left(\frac{z+d}{\sqrt{2\pi}}\right)}{e} \right)}  + \frac{1}{2}$$
with $d = \frac{\sqrt{2\pi}}{e} - \Gamma(1.461632) \approx 0.036534 $ and $W(x)$ is the Lambert W function.
Hence, from $(2)$, we have
$$x \approx f(y):=\frac{\ln \left(\frac{c\Gamma(y)+d}{\sqrt{2\pi}}\right)}{W \left( \frac{\ln \left(\frac{c\Gamma(y)+d}{\sqrt{2\pi}}\right)}{e} \right)}  + \frac{1}{2} -y \tag{3}$$
Remark: You could also use the Stirling's approximation  for $(2)$ and find the relationship of $(x,y)$ as follows
$$(x+y)\ln(x+y) -(x+y) -y\ln(y)+y \approx \ln(c) \tag{4}$$
But the formula $(4)$ is not explicit and besides less accurate than $(3)$.
$$$$
Return back to the formula $(3)$ and the initial problem, it suffices to find the natural numbers $(x,y) \in \Bbb N^2$ such that $$1 \le y \le x \le N$$ with $N$ in the order of $10$ (as mentionned in the question). The algorithm can be as follows

For $y$ from $1$ to $N$
$\qquad$ Calculate $z = f(y)$ with $(3)$
$\qquad$ Stop the loop if $z > N$
$\qquad$ Check if $(x,y)$ satisfies the equation $(2)$ where $x \in \{[z]-K,...,[z]+K\}$
$\qquad$ If any $(x,y)$ is found then $m=x-y-1$

($[z]$ is the entier part of $z$ and $K$ is delibrately fixed but small to prevent the approximation error of $(3)$ , for instant, you can take $K= 2$)
The advantage of this method is the time complexity is $\mathcal{O}(N)$ instead of  $\mathcal{O}(N^2)$ with a naive numerical method (for example, a nested for-loops over $m$ and $n$).
