If \(\frac{^{n}C_{3}}{^{n}P_{2}} = 1\), find the value of n.

Explanation

\(^{n}C_{3} = \frac{n!}{(n - 3)! 3!}\)

\(^{n}P_{2} = \frac{n!}{(n - 2)!}\)

\(\frac{^{n}C_{3}}{^{n}P_{2}} = \frac{n!}{(n - 3)! 3!} ÷ \frac{n!}{(n - 2)!}\)

\(\frac{n!}{(n - 3)! 3!} \times \frac{(n - 2)!}{n!} = \frac{(n - 2)!}{(n - 3)! 3!}\)

Note that \((n - 2)! = (n - 2) \times (n - 2 - 1)! = (n - 2)(n - 3)!\)

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

\(\frac{n - 2}{3!} = 1 \implies n - 2 = 6\)

\(n = 2 + 6 = 8\)