Find the least value of n for which \(^{3n}C_{2} > 0, n \in R\).

  • A \(\frac{1}{3}\)
  • B \(\frac{1}{6}\)
  • C \(\frac{2}{3}\)
  • D 1
waec 2010furthermathematics

The correct answer is C. \(\frac{2}{3}\)

\(^{3n}C_{2} > 0 \implies \frac{3n!}{(3n - 2)! 2!} > 0\)

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

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

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

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

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