WAEC Further Mathematics 2010 Paper

$\binom{3n}{2} > 0 \implies \frac{3n!}{(3n-2)!2!} > 0$

$\frac{3n(3n-1)(3n-2)!}{(3n-2)!2} > 0$

$\frac{3n(3n-1)}{2} > 0$

$3n(3n - 1) > 0 \implies n > 0; n > \frac{1}{3}$

The least number in the option that satisfies $n > 0; n > \frac{1}{3} = \frac{2}{3}$