Equation for those level curves? Related to this question
Let $N\in\mathbb{N}^*,\alpha\in\mathbb{R}$.
What would be the equation $y=f(x)$ for the curve defined by $\ln\binom{N-y}{x}=\alpha$
That's how they look : 

TL;DR : What is the equation $y=f(x)$ for the level curves of $\ln\binom{N-y}{x}$ ?
 A: Problem of turning this to form $y=f(x)$, when the original problem is $f(x,y)=a$ can be done with the following procedure: (in case you can't manipulate the original function to that form)


*

*Take intersection of $f(x,y)=a$ and a line $x=x_0$. You get $f(x_0,y)=a$, rearrange to $f(x_0,y)-a=0$ and run a root-finding algorithm on $g(y)=f(x_0,y)-a$ to find $y_0$. (it might find several roots, so you need to choose one of them, usually the first one the algorithm will find.)

*Collect all resulting $y_0$'s when varying $x_0$: Use $f(x_0)=y_0(x_0).$ This pretty much uses the fact that any function f(x) can be represented as pairs of the form $<x,f(x)>$, which opens as $<x_0,f(x_0)>, <x_1,f(x_1)>, <x_2,f(x_2)>, ... <x_n, f(x_n)>$ and thus you can use the root finding algorithm to find the $f(x_n)$.

*So basically you need to use a result of root-finding algorithm(which is just single number) and build a function from it. It's based on fact that the first step considers $x_0$ as constant, and 2nd step considers it as a variable.


Now this will only give you bunch of numbers, so you don't get nice function like math people usually like to manipulate...
(oops, this solution might require that your function is continuous -- discrete x and y like in this question breaks the root-finding algorithm, since they pretty much assume continuous functions. But at least need to be careful about that aspect -- slower alternative is to go through all the required positions...)
