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

Answer: (B)

$2$

Since, there is a ${}^4{C}_r$ in the expression, then $r\le 4$

$\frac{1}{{}^5{C}_r}+\frac{1}{{}^6{C}_r}=\frac{1}{{}^4{C}_r}$
$\frac{{}^6{C}_r+{}^5{C}_r}{{}^5{C}_r.{}^6{C}_r}=\frac{1}{{}^4{C}_r}$
$\frac{{}^5{C}_r\left(\frac{2\left(6\right)-r}{6-r}\right)}{{}^5{C}_r.{}^6{C}_r}=\frac{1}{{}^4{C}_r}$
$\frac{12-r}{6-r}=\frac{{}^6{C}_r}{{}^4{C}_r}$
$\frac{12-r}{6-r}=\frac{\frac{6!}{(6-r)!}}{\frac{4!}{(4-r)!}}$
$\frac{12-r}{6-r}=\frac{5\times 6}{(6-r)(5-r)}$
$(12-r)(5-r)=30$
$(12-r)(5-r)=10\times 3$
$12-r=10$ or $r=2$.. similarly $5-r=3$ or $r=2$.
Hence $r=2$ is a solution.
Thus the answer is $r=2$.