Suppose A and B are the coefficients of 30^th and 12^th terms respectively in the binomial expansion of ${(1 + x)}^{2n–1}$. If 2A = 5B, then $n$ is equal to:

$\begin{array}{l}\mathrm{A}{=}^{2\mathrm{n}-1}{\mathrm{C}}_{29}\mathrm{B}{=}^{2\mathrm{n}-1}{\mathrm{C}}_{11}\\ 2^{2\mathrm{n}-1}{\mathrm{C}}_{29}=5^{2\mathrm{n}-1}{\mathrm{C}}_{11}\\ 2\frac{(2\mathrm{n}-1)!}{29!(2\mathrm{n}-30)!}=5\frac{(2\mathrm{n}-1)!}{(2\mathrm{n}-12)!11!}\\ \frac{1}{29\dots 12\cdot 5}=\frac{1}{(2\mathrm{n}-12)(2\mathrm{n}-13)\dots (2\mathrm{n}-29)2}\\ \frac{1}{30\cdot 29\dots 12}=\frac{1}{(2\mathrm{n}-12)(2\mathrm{n}-13)\dots (2\mathrm{n}-29)12}\\ 2\mathrm{n}-12=30\\ \mathrm{n}=21\end{array}$