Question

If ²⁰C_r = ²⁰C_r−10, then ¹⁸C_r is equal to
(a) 4896
(b) 816
(c) 1632
(d) nont of these

Answer 1

(b) 816
$r+r-10=20$ [∵${}^n{C}_x{=}^n{C}_y\Rightarrow n=x+y\mathrm{or}x=y$ ]
$$\begin{gathered} \Rightarrow 2r-10=20 \\ \Rightarrow 2r=30 \\ \Rightarrow r=15 \end{gathered}$$

Now,${}^{18}{C}_r{=}^{18}{C}_{15}$
$\therefore {}^{18}{C}_{15}={}^{18}{C}_3$
${\therefore }^{18}{C}_3=\frac{18}{3}\times \frac{17}{2}\times 16=816$