Question

If $P_{(2n+1,n-1)}{:P}_{(2n-1,n)}=3:5$ then find $n$

Answer 1

$P_{(2n+1,n-1)}:P_{(2n-1,n)}=3:5$

$\frac{{}^{2n+1}{P}_{n-1}}{{}^{2n-1}{P}_n}=\frac{3}{5}$

$\frac{(2n+1)!}{(2n+1-n+1)!}\times \frac{(2n-1-n)!}{(2n-1)!}=\frac{3}{5}$

$\frac{(2n+1)\left(2n\right)(2n-1)!\times (n-1)!}{(n+2)!(2n-1)!}=\frac{3}{5}$

$\frac{(2n+1)\left(2n\right)(n-1)!}{(n+2)(n+1)\left(n\right)(n-1)!}=\frac{3}{5}$

$\frac{(2n+1)}{(n+1)(n+2)}=\frac{3}{10}$

$20n+10=3(n^2+3n+2)$

$20n+10=3n^2+9n+6$

$3n^2-11n-4=0$

$3n^2-1n+n-4=0$

$3n(n-4)+1(n-4)=0$

$(3n+1)(n-4)=0$

$n=\frac{-1}{3}$, $n=4$.