Solving equation involving binomial function Solve for $x$ in terms of $i$ and $j$:
$$
\binom{x}{i} = j
$$
where $x$ is Real; $i$ and $j$ are Integers: $x \geqslant i$, $i \geqslant1$, $j \geqslant 0$.
I came across this problem while trying to unrank combinations $\binom{n}{k}$, pp. 51, Algorithm 2.12. I tried the following:
$$
Minimize \left | \binom{x}{i}-j \right | 
$$
$$
s.t. x\geqslant i
$$
A plot of $|\binom{x}{i}-j|$ for $i=20$ and $j=10$ is the following:

I ignore if there is a closed-form for the real solutions of $x$ in terms of $i$ and $j$.

As a partial answer I generated some solutions using CAS.
For $i=1$: $x_1 = j.$
For $i=2$: 
$$x_1 = 1/2+1/2\,\sqrt {1+8\,j},$$
$$x_2 = 1/2-1/2\,\sqrt {1+8\,j}.$$
For $i=3$:
$$x_1 = 1/3\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}+{\frac {1}{\sqrt [3]{
81\,j+3\,\sqrt {729\,{j}^{2}-3}}}}+1,$$
$$x_2 = -1/6\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-1/2\,{\frac {1}{
\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}}}+1+1/2\,i\sqrt {3} \left( 
1/3\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-{\frac {1}{\sqrt [3]{
81\,j+3\,\sqrt {729\,{j}^{2}-3}}}} \right),$$
$$x_3 = -1/6\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-1/2\,{\frac {1}{
\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}}}+1-1/2\,i\sqrt {3} \left( 
1/3\,\sqrt [3]{81\,j+3\,\sqrt {729\,{j}^{2}-3}}-{\frac {1}{\sqrt [3]{
81\,j+3\,\sqrt {729\,{j}^{2}-3}}}} \right).$$
For $i=4$:
$$x_1 = 3/2+1/2\,\sqrt {5+4\,\sqrt {1+24\,j}},$$
$$x_2 = 3/2-1/2\,\sqrt {5+4\,\sqrt {1+24\,j}},$$
$$x_3 = 3/2+1/2\,\sqrt {5-4\,\sqrt {1+24\,j}},$$
$$x_4 = 3/2-1/2\,\sqrt {5-4\,\sqrt {1+24\,j}}.$$
For $i=5$ I didn't find closed-form.
And the solutions for $i=6$ are here. 
I'm also very interested in a general solution.
