Question

If ⁿC_{r – 1} = 36, ⁿC_r = 84 and ⁿC_{r + 1} = 126, then r = ____________.

Answer 1

nC_r-1 = 36, ⁿC_r = 84 and ⁿC_{r + 1} = 126
$$\begin{gathered} \mathrm{here}\frac{{}^{n}C_r}{{}^{n}C_{r-1}}=\frac{84}{36} \\ \mathrm{i}.\mathrm{e}\frac{n!}{r!\left(n-r\right)!}\times \frac{\left(r-1\right)!\left(n-r+1\right)!}{n!}=\frac{7}{3} \\ \mathrm{i}.\mathrm{e}\frac{\left(r-1\right)!\left(n-r+1\right)\left(n-r\right)!}{r\left(r-1\right)!\left(n-r\right)!}=\frac{7}{3} \\ \mathrm{i}.\mathrm{e}\frac{n-r+1}{r}=\frac{7}{3} \\ \mathrm{i}.\mathrm{e}3n-3r+3=7r \\ \mathrm{i}.\mathrm{e}3n+3=10r\left(1\right) \\ \mathrm{also},\frac{{}^{n}C_{r+1}}{{}^{n}C_r}=\frac{126}{84} \\ \frac{n!}{\left(r+1\right)!\left(n-r-1\right)!}\times \frac{r!\left(n-r\right)!}{n!}=\frac{3}{2} \\ \mathrm{i}.\mathrm{e}\frac{r!\left(n-r\right)\left(n-r-1\right)!}{\left(r+1\right)r!\left(n-r-1\right)!}=\frac{3}{2} \end{gathered}$$