If frac{{}^{n + 2}{C_6}}{{}^{n - 2}{P_2}} = 1 | vedclass

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(C) Given the equation: $\frac{{}^{n + 2}{C_6}}{{}^{n - 2}{P_2}} = 11$Using the formulas ${}^nC_r = \frac{n!}{r!(n-r)!}$ and ${}^nP_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 ${}^nC_r = \frac{n!}{r!(n-r)!}$ and ${}^nP_r = \frac{n!}{(n-r)!}$:$\frac{\frac{(n+2)!}{6!(n-4)!}}{\frac{(n-2)!}{(n-4)!}} = 11$$\frac{(n+2)!}{6! \cdot (n-2)!} = 11$$\frac{(n+2)(n+1)(n)(n-1)(n-2)!}{720 \cdot (n-2)!} = 11$$(n+2)(n+1)(n)(n-1) = 11 \cdot 720 = 7920$Testing integer values for $n$,if $n=9$:$(11)(10)(9)(8) = 7920$. Thus,$n=9$ is the solution.Now,check which equation is satisfied by $n=9$:For option $C$: $n^2 + 3n - 108 = (9)^2 + 3(9) - 108 = 81 + 27 - 108 = 108 - 108 = 0$.Therefore,$n$ satisfies the equation $n^2 + 3n - 108 = 0$.

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

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