Let's say we have an inequality, $p \le {a \choose b}$ where $p$ is a fixed constant and $a, b$ are variables. The problem is that, we are trying to find the minimum $a$ with respect to the inequality $p \le {a \choose b}$. Is there a closed form solution (can be approximate as well/doesn't have to be exact) for that combinatorial optimization problem? 
Solution for a combinatorial minimization problem
optimization;combinatorics
G. Bach noticed that the best choice for $b$ is $\lfloor a/2 \rfloor$, and furthermore this choice gives the maximal binomial coefficient. The easiest way to see this is to consider the ratio of two adjacent binomial coefficients:$$ \frac{\binom{a}{b}}{\binom{a}{b+1}} = \frac{b+1}{a-b}.$$Therefore $\binom{a}{b} \leq \binom{a}{b+1}$ iff $b+1 \leq a-b$ iff $2b+1 \leq a$, and we get the desired result.The binomial theorem states that$$ \sum_{b=0}^a \binom{a}{b} = 2^a, $$hence$$ \frac{2^a}{a+1} \leq \binom{a}{\lfloor a/2 \rfloor} \leq 2^a. $$(The correct order of magnitude, $\Theta(2^a/\sqrt{a})$, can be found using Stirling's approximation.) Therefore the optimal $a$ satisfies the following inequalities:$$ \frac{2^{a-1}}{a} < p \leq 2^a. $$Therefore $a \geq \log_2 p$. On the other hand, when $p \geq 4$, $a \geq 2$, and so $a - \log_2 a \geq a/2$ (since $a-\log_2 a$ is monotone and $2 - \log_2 2 = 2/2$). Therefore for $p \geq 4$,$$ 2p\log_2 (2p) > 2\frac{2^{a-1}}{a} (a - \log_2 a) > 2^{a-1}.$$We conclude that$$ \log_2 p \leq a < \log_2 p + 1 + \log_2 \log_2 (2p). $$Therefore the binary search that G. Bach mentions takes only $O(\log\log\log p)$ steps.If all we want is an asymptotic expression, then since $\binom{a}{\lfloor a/2 \rfloor} = \Theta(2^a/\sqrt{a})$, approximately we have $2^a/\sqrt{a} = Cp$, and so $a = \log_2 p + \Theta(\log_2\log_2 p)$. More terms can be obtained by taking more terms in Stirling's approximation. For example, we can find a constant $A$ such that $a = \log_2 p + A\log_2\log_2 p + o(\log_2\log_2 p)$.