Question

If ${}^{n+2}{C}_8:{}^{n-2}{P}_4=57:16,$ then the value of n is:

• A. 20
• B. 19
• C. 18
• D. 17

Answer 1

Answer: (B)

Given ${}^{n+2}{C}_8{:}^{n-2}{P}_4=57:16$
$\frac{{}^{n+2}{C}_8}{{}^{n-2}{P}_4}=\frac{57}{16}\left[{\because }^n{C}_r=\frac{n!}{r!\left(n-r\right)!}{\text{and}}^n{P}_r=\frac{n!}{\left(n-r\right)!}\right]$
$\Rightarrow \frac{\left(n+2\right)!}{8!\left(n+2-8\right)!}\times \frac{\left(n-2-4\right)!}{\left(n-2\right)!}=\frac{57}{16}$
$\Rightarrow \frac{\left(n+2\right)\left(n+1\right)n.\left(n-1\right)}{\mathrm{8.7.6.5.4.3.2.1}}=\frac{57}{16}$
$\Rightarrow \left(n+2\right)\left(n+1\right)n\left(n-1\right)=143640$
$\Rightarrow \left(n^2+n-2\right)\left(n2+n\right)=143640$
$\Rightarrow {\left(n^2+n\right)}^2-2\left(n^2+n\right)+1=143640+1$
$\Rightarrow {\left(n^2+n-1\right)}^2={\left(379\right)}^2$
$\Rightarrow n^2+n-1=379\left[\because n^2+n-1>0\right]$
$\Rightarrow n^2+n-1-379=0$
$\Rightarrow n^2+n-380=0$
$\Rightarrow \left(n+20\right)\left(n-19\right)=0$
$\Rightarrow n=-20,n=19$
$\because$ n is not negative.
$\therefore n=19$