If $^{n} C_{r - 1} = 36,^{n} C_{r} = 84$ and $^{n} C_{r + 1} = 126$ then $^{r} C_{2} =$

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answer is B.

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Detailed Solution

$\frac{n_{C_{r}}}{n_{C_{r - 4}}} = \frac{n - r + 1}{r}$

$\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{84}{36} \begin{aligned} \Rightarrow \frac{n - r + 1}{r} = \frac{7}{3} \\ \Rightarrow 3 n - 3 r + 3 = 7 r \\ \Rightarrow 10 r = 3 n + 3 - \dots \end{aligned}$

$\begin{aligned} \frac{^{n} C_{r + 1}}{^{n} C_{r}} = \frac{126}{84} \\ \Rightarrow \frac{\frac{n!}{(r + 1)! (n - r - 1)!}}{n!} = \frac{3}{2} \\ \Rightarrow \frac{n - r}{(r)! (n - r)!} = \frac{3}{2} \\ \Rightarrow 2 n - 2 r = 3 r + 3 \\ \Rightarrow 5 r = 2 n - 3 - \dots \end{aligned}$

$\begin{aligned} \begin{aligned} \frac{10 r}{5 r} = \frac{3 n + 3}{2 n - 3} \\ \Rightarrow 2 (2 n - 3) = 3 n + 3 \\ \Rightarrow 4 n - 6 = 3 n + 3 \\ \Rightarrow n = 6 + 3 \\ \Rightarrow n = 9 \end{aligned} \\ \begin{aligned} 10 r = 3 n + 3 \\ \Rightarrow 3 (9) + 3 = 30 \\ \Rightarrow 10 r = 30 \\ \Rightarrow r = 3 \end{aligned} \\ ∴ n = 9, r = 3 \end{aligned}$