Question

Find r if

(i) ⁵P_r = 2⁶P_r-1 

(ii) ⁵P_r = ⁶P_r-1

Answer 1

(i) ⁵P_r = 2⁶P_r-1 

\(⇒\frac{5!}{\left(5-r\right)!}=2\times\frac{6!}{\left(6-r+1\right)!}\)

\(⇒\frac{5!}{\left(5-r\right)!}=2\times\frac{6!}{\left(7-r\right)!}\)

\(⇒\frac{5!}{\left(5-r\right)!}=\frac{2\times6\times5!}{\left(7-r\right)\left(6-r\right)\left(5-r\right)!}\)

\(⇒1=\frac{2\times6}{\left(7-r\right)\left(6-r\right)}\)
\(⇒\left(7-r\right)\left(6-r\right)=12\)
\(⇒42-6r-7r+r^2=12\)
\(⇒r^2-13r+30=0\)
\(⇒r^2-3r-10r+30=0\)
\(⇒r\left(r-3\right)-10\left(r-3\right)=0\)
\(⇒\left(r-3\right)\left(r-10\right)=0\)
\(⇒\left(r-3\right)=0\space or \left(r-10\right)=0\)
\(⇒r=3 \) or \( r=10\)
\(^nP_r=\frac{n!}{\left(n-r\right)!}\), where \(0≤r≤n\)
It is known that, 
\(∴0 ≤ r ≤ 5 \)
Hence, \(r \neq 10 ∴r = 3\)

 

(ii) ⁵P_r = ⁶P_r-1

\(⇒\frac{5!}{\left(5-r\right)!}=\frac{6!}{\left(6-r+1\right)!}\)

\(⇒\frac{5!}{\left(5-r\right)!}=6\times\frac{5!}{\left(7-r\right)!}\)

\(⇒\frac{1}{\left(5-r\right)!}=\frac{6}{\left(7-r\right)\left(6-r\right)\left(5-r\right)!}\)

\(⇒1=\frac{6}{\left(7-r\right)\left(6-r\right)}\)
\(⇒\left(7-r\right)\left(6-r\right)=6\)
\(⇒42-7r-6r+r^2-6=0\)
\(⇒r^2-13r+36=0\)
\(⇒r^2-4r-9r+36=0\)
\(⇒r\left(r-4\right)-9\left(r-4\right)=0\)
\(⇒\left(r-4\right)\left(r-9\right)=0\)
\(⇒\left(r-4\right)=0\) or \(\left(r-9\right)=0\)
\(⇒r=4\) or \(r=9\)

\(^nP_r=\frac{n!}{\left(n-r\right)!}\), where\( 0≤r≤n\)
It is known that,
\(0 ≤ r ≤ 5\)
Hence, \(r \neq 9 \)
\(∴ r = 4\)