Q.7.    Find r if

          (i) ^{5}P_{r}=2\; ^{6}\! P_{r -1}

Answers (1)
S seema garhwal

Given : ^{5}P_{r}=2\; ^{6}\! P_{r -1}

To find the value of r.

^{5}P_{r}=2\; ^{6}\! P_{r -1}

\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-(r-1))!}

\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6!}{(6-r+1)!}

\Rightarrow \, \, \frac{5!}{(5-r)!}=2\times \frac{6\times 5!}{(6-r+1)!}

\Rightarrow \, \, \frac{1}{(5-r)!}=2\times \frac{6}{(7-r)!}

\Rightarrow \, \, \frac{1}{(5-r)!}= \frac{12}{(7-r)\times (6-r)\times (5-r)!}

\Rightarrow \, \, \frac{1}{1}= \frac{12}{(7-r)\times (6-r)}

\Rightarrow \, \, (7-r)\times (6-r)=12

\Rightarrow \, \, 42-6r-7r+r^2=12

\Rightarrow \, \, r^2-13r+30=0

\Rightarrow \, \, r^2-3r-10r+30=0

\Rightarrow \, \, r(r-3)-10(r-3)=0

\Rightarrow \, \, (r-3)(r-10)=0

\Rightarrow \, \, r=3,10

We know that 

                           ^nP_r=\frac{n!}{(n-r)!}       

                                   where 0\leq r\leq n

                           \therefore 0\leq r\leq 5

Thus the  value of, r=3

                           

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Exams
Articles
Questions