Question

Find n if ^n-1P₃: ⁿP₃ = 1: 9.

Answer 1

^n-1P₃: ⁿP₃ = 1: 9

\(⇒\frac{^{n-1}P_3}{^nP_3} = \frac{1}{ 9}\)

\(⇒\frac{\left[\frac{(n-1)!}{(n-1-3)!}\right]}{\left[\frac{n!}{(n-4)!}\right]}=\frac{1}{9}\)

\(⇒\frac{(n-1)!}{(n-4)!}\times\frac{(n-4)!}{n!}=\frac{1}{9}\)

\(⇒\frac{(n-1)!}{n\times(n-1)!}=\frac{1}{9}\)

\(⇒\frac{1}{n}=\frac{1}{9}\)

\(∴n=9\)