Thank you for reporting, we will resolve it shortly
Q.
If ${ }^{2 n+1} P _{n-1}:{ }^{2 n-1} P _n=11: 21$, then $n^2+n+15$ is equal to :
JEE MainJEE Main 2023Permutations and Combinations
Solution:
$ \frac{(2 n +1) !( n -1) !}{( n +2) !(2 n -1) !}=\frac{11}{21} $
$ \Rightarrow \frac{(2 n +1)(2 n )}{( n +2)( n +1) n }=\frac{11}{21} $
$ \Rightarrow \frac{2 n +1}{( n +1)( n +2)}=\frac{11}{42} $
$ \Rightarrow n =5 $
$ \Rightarrow n ^2+ n +15$
$=25+5+15=45$