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Solution:

Given: ${ }^{20} C_{r}={ }^{20} C_{r+6}$

To find: $r=?$

We know that:

${ }^{n} C_{r}={ }^{n} C_{n-r}$

$\Rightarrow{ }^{20} \mathrm{C}_{\mathrm{r}+6}={ }^{20} \mathrm{C}_{20-(\mathrm{r}+6)}$

$\Rightarrow{ }^{20} \mathrm{C}_{\mathrm{r}+6}={ }^{20} \mathrm{C}_{20-\mathrm{r}-6}={ }^{20} \mathrm{C}_{14-\mathrm{r}}$

$\Rightarrow{ }^{20} \mathrm{C}_{14-\mathrm{r}}={ }^{20} \mathrm{C}_{\mathrm{r}}$

$\Rightarrow 14-r=r$

$\Rightarrow 2 r=14$

⇒ $r=\frac{14}{2}=7$

Ans:r=7

(ii) Given: ${ }^{18} \mathrm{Cr}={ }^{18} \mathrm{C}_{\mathrm{r}+2}$

To find: ${ }^{r} C_{5}=?$

We know that:

${ }^{n} C_{r}={ }^{n} C_{n-r}$

$\Rightarrow{ }^{18} \mathrm{C}_{\mathrm{r}+2}={ }^{18} \mathrm{C}_{18-(\mathrm{r}+2)}$

$\Rightarrow{ }^{18} \mathrm{C}_{\mathrm{r}+2}={ }^{18} \mathrm{C}_{18-\mathrm{r}-2}={ }^{18} \mathrm{C}_{16-\mathrm{r}}$

$\Rightarrow{ }^{18} \mathrm{C}_{16-\mathrm{r}}={ }^{18} \mathrm{C}_{\mathrm{r}}$

$\Rightarrow 16-\mathrm{r}=\mathrm{r}$

$\Rightarrow 2 \mathrm{r}=16$

⇒  $r=\frac{16}{2}=8$

So,

${ }^{\mathrm{r}} \mathrm{C}_{5}={ }^{8} \mathrm{C}_{5}$

$\Rightarrow{ }^{8} C_{5}=\frac{8 !}{(8-5) ! \times 5 !}$

$\Rightarrow{ }^{8} C_{5}=\frac{8 !}{3 ! \times 5 !}$

$\Rightarrow{ }^{8} \mathrm{C}_{5}=\frac{8 \times 7 \times 6 \times 5 !}{3 ! \times 5 !}$

$\Rightarrow{ }^{8} \mathrm{C}_{5}=\frac{8 \times 7 \times 6}{3 !}$

$\Rightarrow{ }^{8} \mathrm{C}_{5}=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}$

$\Rightarrow{ }^{8} \mathrm{C}_{5}=56$

Ans: ${ }^{8} \mathrm{C}_{5}=56$