Solve the following
Question:

If 20Cr = 20Cr + 4 , then rC3 is equal to

(a) 54

(b) 56

(c) 58

(d) none of these

Solution:

(b) 56

$r+r+4=20 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$

$\Rightarrow 2 r+4=20$

$\Rightarrow 2 r=16$

$\Rightarrow r=8$

Now, ${ }^{r} C_{3}={ }^{8} C_{3}$

${ }^{8} C_{3}=\frac{8 !}{3 ! 5 !}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56$

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