Tag: factorial notation

Consider the given value of permutation,$^{n}{{P} _{r-1}}{{:}^{15}}{{P} _{r-2}}=3:4$

  $ \,\,\,\,\,\,\dfrac{^{15}{{P} _{r-1}}}{^{15}{{P} _{r-2}}}=\dfrac{3}{4} $

 $ \,\,\,\dfrac{\dfrac{15!}{(15-r+1)!}}{\dfrac{15!}{\left( 15-r+2 \right)!}}=\dfrac{3}{4} $

 $  $

 $ \,\,\,\,\,\,\dfrac{\left( 15-r+2 \right)!}{(15-r+1)!}=\dfrac{3}{4} $

 $ \,\,\,\,\,\dfrac{\left( 17-r \right)!}{(16-r)!}=\dfrac{3}{4} $

 $ \,\,\,\dfrac{\left( -17+r \right)!}{(-16+r)!}=\dfrac{3}{4} $

 $ \dfrac{\left( r-17 \right)!}{(r-16)!}=\dfrac{3}{4} $

 $ \dfrac{\left( r-17 \right)\left( r-16 \right)!}{(r-16)!}=\dfrac{3}{4} $

 $ r-17=\dfrac{3}{4} $

 $ r=\dfrac{71}{4} $

This is the required answer.

Given $T _n=\ ^nC _3$

  Given $^{2n+1}P _{n-1}: ^{2n-1}P _n = 3 : 5$ which can be further simplified as
$ \displaystyle \frac {\frac { (2n+1)! }{ (n+2)! }  }{ \frac { (2n-1)! }{ (n-1)! }  }= \frac { 3 }{ 5 } $
$\Rightarrow \displaystyle \frac {(2n+1)(2n)}{(n+2)(n+1)(n)} = \displaystyle\frac{3}{5} $
$\Rightarrow $ $3{n}^{2}-11n-4=0$ .
On solving this quadratic equation we get roots as  $ 4 $ and $ -\displaystyle \frac{1}{3} $.
Since $n$ is an integer,$n=4$.