Question

${}^n{P}_7=42\times {}^n{P}_5$ then $n=$

• A. $5$
• B. $-1$
• C. $7$
• D. $12$

Answer 1

Answer: (D)

$12$

$\frac{{}^n{P}_7}{{}^n{P}_5}=42$
$\frac{\frac{n!}{(n-7)!}}{\frac{n!}{(n-5)!}}=42$
$\Rightarrow (n-5)(n-6)=42$
$\Rightarrow n^2-11n+30=42$
$\Rightarrow n^2-11n-12=0$
$\Rightarrow n^2-12n+n-12=0$
$\Rightarrow n(n-12)+1(n-12)=0$
$\Rightarrow (n-12)(n+1)=0$
$n=12,-1$
$\Rightarrow n=12$
($\because n$ can not be negative)