Question

If $\frac{1}{{}^5{C}_r}+\frac{1}{{}^6{C}_r}=\frac{1}{{}^4{C}_r},$ then the value of $r$ equals to

• A. $4$
• B. $2$
• C. $5$
• D. $3$

Answer 1

The correct option is B $2$

$\frac{1}{{}^5{C}_r}+\frac{1}{{}^6{C}_r}=\frac{1}{{}^4{C}_r}\Rightarrow \frac{r!(5-r)!}{5!}+\frac{r!(6-r)!}{6!}=\frac{r!(4-r)!}{4!}\frac{(5-r)}{5}+\frac{(6-r)\times (5-r)}{6\times 5}=1\Rightarrow r=2$