Answer: C
Explanation:
We know that the maximum nodes in a binary tree of height, $\color {red}{\mathrm h = 2^{\mathrm h + 1} - 1} $ nodes.
Also, for $\mathrm n$ elements the decision would be taken from $\color {red}{\mathrm n!}$ elements.
$\Rightarrow 2^{\mathrm h+1} - 1 \ge\mathrm n!\\\Rightarrow \mathrm h \ge \log_2 \mathrm n! + 1\\\Rightarrow \mathrm h \ge \log_2\mathrm n!\\\Rightarrow \mathrm h \ge\mathrm n\log\mathrm n \;\;\;\;\;\;\text{[Using Stirling's approximation, $\mathrm n! = \frac{(\mathrm n^{\mathrm n})}{e}]$}\\\Rightarrow \Omega( \mathrm n \log\mathrm n)$
$\therefore \; \textbf C$ is the correct option.