If frac{{{}^{n + 2}{Ct}}}{{{}^{n - 2}{P₂}}} = 11 , then n satisfies equation

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6.Permutation and Combination

hard

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

A

$n^2 + n - 110 =0$

B

$n^2 + 2n - 80 =0$

C

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

D

$n^2 + 5n - 84 =0$

(JEE MAIN-2016)

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}}}}{{\frac{{(n – 2)(n – 3)}}{{2.1}}}} = 11$

$ \Rightarrow (n + 2)(n + 1)n(n – 1) = 11.10.9.4$

$ \Rightarrow n = 9$

${n^2} + 3n – 108 = {(9)^2} + 3(9) – 108$

$ = 81 + 27 – 108$

$ = 108 – 108 = 0$

Standard 11

Mathematics

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