If frac{{}^{n + 2}C_6}{{}^{n - 2}P_2} = 11,th | vedclass

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(C) Given the equation: $\frac{{}^{n + 2}C_6}{{}^{n - 2}P_2} = 11$Using the formulas ${}^{n}C_r = \frac{n!}{r!(n-r)!}$ and ${}^{n}P_r = \frac{n!}{(n-r

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

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(C) Given the equation: $\frac{{}^{n + 2}C_6}{{}^{n - 2}P_2} = 11$Using the formulas ${}^{n}C_r = \frac{n!}{r!(n-r)!}$ and ${}^{n}P_r = \frac{n!}{(n-r)!}$:$\frac{\frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{6 	imes 5 	imes 4 	imes 3 	imes 2 	imes 1}}{\frac{(n-2)!}{(n-2-2)!}} = 11$$\frac{(n+2)(n+1)n(n-1)(n-2)(n-3)}{720} 	imes \frac{1}{(n-2)(n-3)} = 11$$(n+2)(n+1)n(n-1) = 11 	imes 720 = 7920$Testing values for $n$,if $n=9$:$(11)(10)(9)(8) = 7920$Thus,$n=9$ is the solution.Checking the options for $n=9$:For option $C$: $n^2 + 3n - 108 = (9)^2 + 3(9) - 108 = 81 + 27 - 108 = 0$.Therefore,$n$ satisfies $n^2 + 3n - 108 = 0$.

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

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