JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 2)
If $${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$$ = 11, then n satisfies the
equation :
n2 + 3n − 108 = 0
n2 + 5n − 84 = 0
n2 + 2n − 80 = 0
n2 + n − 110 = 0
Explanation
$${{{}^{n + 2}{C_6}} \over {{}^{n - 2}{P_2}}} = 11$$
$$ \Rightarrow $$ $${{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}$$
$$ \Rightarrow $$ (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 . 6 . 5 . 4 . 3 . 2 . 1
$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 10 . 9 . 8
$$ \therefore $$ n = 9
This value of n satisfy the equation,
n2 + 3n $$-$$ 108 = 0
$$ \Rightarrow $$ $${{\left( {n + 2} \right)!} \over {6!\,\left( {n - 4} \right)!}} = 11\,.\,{{\left( {n - 2} \right)!} \over {\left( {n - 4} \right)!}}$$
$$ \Rightarrow $$ (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 . 6 . 5 . 4 . 3 . 2 . 1
$$ \Rightarrow $$ (n + 2) (n + 1) n (n $$-$$ 1) = 11 . 10 . 9 . 8
$$ \therefore $$ n = 9
This value of n satisfy the equation,
n2 + 3n $$-$$ 108 = 0
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