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.

  • 1
    $\begingroup$ 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? $\endgroup$
    – JMoravitz
    Nov 30, 2023 at 17:20
  • $\begingroup$ One can extend to non-integer $n$ and $k$ using the beta function. $\endgroup$
    – coiso
    Nov 30, 2023 at 17:49
  • $\begingroup$ @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? $\endgroup$ Nov 30, 2023 at 19:15
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 1, 2023 at 18:07
  • $\begingroup$ @ТymaGaidash Thanks. I don't think i can follow that yet. If you have any resources you can refer me to would be great. $\endgroup$ Dec 1, 2023 at 18:53


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