Solve the following

Question:

If 16Cr = 16Cr + 2, find rC4.

Solution:

Given;

${ }^{16} C_{r}={ }^{16} C_{r+2}$

$16=r+r+2 \quad\left[\because\right.$ Property $5:{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow x=y$ or $\left.x+y=n\right]$

$\Rightarrow 2 r+2=16$

$\Rightarrow 2 r=14$

$\Rightarrow r=7$

Now, ${ }^{r} C_{4}={ }^{7} C_{4}$

$\Rightarrow{ }^{7} C_{4}={ }^{7} C_{3} \quad\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$

$\Rightarrow{ }^{7} C_{4}={ }^{7} C_{3}=\frac{7}{3} \times \frac{6}{2} \times \frac{5}{1} \times{ }^{4} C_{0} \quad\left[\because{ }^{n} C_{r}=\frac{n}{r} \cdot{ }^{n-1} C_{r-1}\right]$

$\Rightarrow^{7} C_{4}=35 \quad\left[\because{ }^{n} C_{0}=1\right]$

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