If \(^nC_2\) = 15, find the value of n
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
\(^nC_2\) = 15
\(\frac{n!}{(n - 3)! 2!}\) = 15
\(\frac{n(n - 1)(n - 2)!}{(n - 2)!2}\) = 15
n\(^2\) - n = 30
n\(^2\) - n - 30 = 0
(n - 6)(n + 5) = 0
n = 6 or n = -5
\(^nC_2\) = 15
\(\frac{n!}{(n - 3)! 2!}\) = 15
\(\frac{n(n - 1)(n - 2)!}{(n - 2)!2}\) = 15
n\(^2\) - n = 30
n\(^2\) - n - 30 = 0
(n - 6)(n + 5) = 0
n = 6 or n = -5