Q53 / 90 · JEE Main 2016 (10 Apr Online) · Mathematics

If $\frac{^{n + 2} C_{6}}{^{n - 2} P_{2}} = 11$ , then $n$ satisfies the equation:

A
$n^{2} + n - 110 = 0$
B
$n^{2} + 2 n - 80 = 0$
Correct answer
C
$n^{2} + 3 n - 108 = 0$
D
$n^{2} + 5 n - 84 = 0$

Solution

\frac{^{n + 2} C_{6}}{^{n - 2} P_{2}} = 11

\Rightarrow

\frac{\frac{(n + 2) (n + 1) n (n - 1) (n - 2) (n - 3)}{6 . 5 . 4 . 3 . 2 . 1}}{(n - 2) (n - 3)} = 11

\Rightarrow

(n + 2) (n + 1) n (n - 1) = 11.10.9 . 8

\Rightarrow

n = 9

Hence, the correct answer is option: 'C'.

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