Permutation and Combinations — MATHS STD 11 Science — Question

2. 56

Solution:

$\text{r}+\text{r}+4=20$

$\Rightarrow 2\text{r}+4=20$

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

$\Rightarrow \text{r}=8$

Now,

${^\text{r}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$

$\therefore\ {^\text{8}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$

$\therefore\ {^\text{8}}\text{C}_{\text{3}}=\frac{8!}{3!5!}$

$=\frac{8\times7\times6}{3\times2\times1}=56$