If frac{{{}^{n + 2}{C_6}}}{{{}^{n - 2}{P | vedclass

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Question

$\frac{{^{n + 2}{C_6}}}{{^{n - 2}{P_2}}} = 11$

$ \Rightarrow \frac{{\frac{{(n + 2)(n + 1)n(n - 1)(n - 2)(n - 3)}}{{6.5.4.3.2.1}}}}{{\frac{{(n - 2)(

Vedclass · Accepted Answer

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

Vedclass · Answer

$\frac{{^{n + 2}{C_6}}}{{^{n - 2}{P_2}}} = 11$

$ \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$

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

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