Question

If $\frac{^{n+2}C_6}{^{n-2}P_2} = 11$, then $n$ satisfies the equation :

• A. $n^2 + 3n - 108 =0$
• B. $n^2 + 5n - 84 =0$
• C. $n^2 + 2n - 80 =0$
• D. $n^2 + n - 110 =0$

Answer 1

Answer: (A)

$n^2 + 3n - 108 =0$

Answer 2

$\frac{{ }^{ n +2} C _{6}}{ n -2 P _{2}} =11$
$\Rightarrow \frac{( n +2) !}{6 !( n -4) !}=11 .\frac{( n -2) !}{( n -4) !}$
$\Rightarrow ( n +2) !=11 . 6 !( n -2) !$
$\Rightarrow ( n +2)( n +1) n ( n -1)=11.6 !$
$\Rightarrow ( n +2)( n +1) n ( n -1)=11.10 .9 .8$
$\Rightarrow n +2=11$
$\Rightarrow n =9$
Which satifies the $n^{2}+3 n-108=0$