Question

If 2 × ⁿC₅ = 9 × ^{n – 2}C₅, then n = ____________.

Answer 1

$$\begin{gathered} 2\times {}^{n}C_5=9\times {}^{n-2}C_5 \\ \mathrm{i}.\mathrm{e}2\times \frac{n!}{5!\left(n-5\right)!}=9\times \frac{\left(n-2\right)!}{5!\left(n-2-5\right)!} \\ \mathrm{i}.\mathrm{e}\frac{2\times n!}{5!\left(n-5\right)!}=\frac{9\times \left(n-2\right)!}{5!\left(n-7\right)!} \\ \mathrm{i}.\mathrm{e}\frac{2\times n\left(n-1\right)\left(n-2\right)!}{5!\left(n-5\right)\left(n-6\right)\left(n-7\right)!}=\frac{9\times \left(n-2\right)!}{5!\left(n-7\right)!} \\ \mathrm{i}.\mathrm{e}2\times \left[n\left(n-1\right)\right]=9\left(n-5\right)\left(n-6\right) \\ \mathrm{i}.\mathrm{e}2n^2-2n=9\left[n^2-11n+30\right] \\ 2n^2-2n=9n^2-99n+270 \\ \mathrm{i}.\mathrm{e}7n^2-97n+270 \\ D=9409-4\left(7\right)\left(270\right) \\ =9409-7560=1849 \\ \mathrm{Using},\mathrm{formula}\frac{-b\pm \sqrt{D}}{2a}=\frac{97\pm \sqrt{1849}}{2\times 7}=\frac{97\pm 43}{14} \\ =10,\frac{27}{7} \\ \mathrm{Since}n\in N,\Rightarrow n=10 \end{gathered}$$