Root of the polynomial $x(x-1)(x-2)\cdots(x-K)=C$

Question

Is there an analytic way to obtain the highest root of the polynomial $x(x-1)(x-2)\cdots(x-K)=C$ in terms of $K$ and $C$? The integer $K \ll x$ and the constant $C$ are known.

The other way to ask the question is given $K$ and $C$ find $N$ such that $\binom{N}{K}=C$

Answer 1

Doesn't Stirling formula help? Doing some simplifications using $K \ll N$, the question becomes something like finding a solution to $$ N \log N + (K-N) \log(N-K) = K + \log C $$ (plus a small error term like $\frac{1}{2} \log\frac{N}{N-K}$). Depending on how small $K$ is with respect to $N$, maybe you can also ignore the difference between $\log(N)$ and $\log(N-K)$ in the equation above and find an approximated solution explicitly.

Answer 2

Is there an analytic way to obtain the highest root of the polynomial $x(x-1)\cdots(x-K)\!=\!C$

Doubtful, since it is well-known that there is no general formula for the roots of a polynomial with degree greater than $4$. Also, the inverse of the factorial or falling factorial function is not known to possess a closed form.