Solveeit Logo
Login

Question

Mathematics Question on Permutations

Find n if n-1P3: nP3 = 1: 9.

Answer

n-1P3: nP3 = 1: 9

⇒n−1P3nP3=19⇒\frac{^{n-1}P_3}{^nP_3} = \frac{1}{ 9}

⇒[(n−1)!(n−1−3)!][n!(n−4)!]=19⇒\frac{\left[\frac{(n-1)!}{(n-1-3)!}\right]}{\left[\frac{n!}{(n-4)!}\right]}=\frac{1}{9}

⇒(n−1)!(n−4)!×(n−4)!n!=19⇒\frac{(n-1)!}{(n-4)!}\times\frac{(n-4)!}{n!}=\frac{1}{9}

⇒(n−1)!n×(n−1)!=19⇒\frac{(n-1)!}{n\times(n-1)!}=\frac{1}{9}

⇒1n=19⇒\frac{1}{n}=\frac{1}{9}

∴n=9∴n=9