Solve the following


If 20Cr = 20Cr−10, then 18Cr is equal to

(a) 4896

(b) 816

(c) 1632

(d) nont of these


(b) 816

$r+r-10=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$

$\Rightarrow 2 r-10=20$

$\Rightarrow 2 r=30$

$\Rightarrow r=15$

Now, ${ }^{18} C_{r}={ }^{18} C_{15}$

$\therefore{ }^{18} C_{15}={ }^{18} C_{3}$

$\therefore{ }^{18} C_{3}=\frac{18}{3} \times \frac{17}{2} \times 16=816$

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