# Looking for a function to solve for $k$ in a binomial coefficient. $n \choose k$

Looking for a straightforward function $$f(n,a) = k$$. In other words solve for $$k$$ in $${n \choose k} = a$$ for any given $$n$$ and $$a$$.

If it could be valid for non-natural values of $$k$$ as well that would be wondeful. (I'm missing the background knowledge of the gamma function to know if this even possible).

Thank you.

• So, $f(n,a)=k$ whenever $\binom{n}{k}=a$... what about when there are no such $k$? What should be $f(4,2)$ noting that $\binom{4}{0}=1,\binom{4}{1}=4,\dots$ and noting that the generalized binomial coefficient only allows for the top number to be non-natural but the bottom number must always be natural? Nov 30, 2023 at 17:20
• One can extend to non-integer $n$ and $k$ using the beta function. Nov 30, 2023 at 17:49
• @JMoravitz If we simply envision $n \choose k$ as carrying out the function of $n!$, $k$ amount of times, instead of a literal binomial coeffecient, is it not possible for it to be generalized for non-natural numbers? Nov 30, 2023 at 19:15
• You would need to invert $\binom nk=(-1)^n\frac{n!\sin(\pi k)}{\pi \prod\limits_{j=0}^n (k-j)},n\in\Bbb N$. It is fairly difficult to invert $\binom 0k=\frac{\sin(\pi k)}{\pi k}$ already. For non-natural $n$, you would need the inverse of a gamma like function. Furthermore, there are infinite branches for the inverse. Dec 1, 2023 at 18:07
• @ТymaGaidash Thanks. I don't think i can follow that yet. If you have any resources you can refer me to would be great. Dec 1, 2023 at 18:53