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If $\frac{1}{{ }^{5} C_{r}}+\frac{1}{{ }^{6} C_{r}}=\frac{1}{{ }^{4} C_{r}},$ then the value of $r$ is
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2
Given, $\frac{1}{5 C_{r}} +\frac{1}{6 C_{r}} =\frac{1}{4 C_{r}}$
$\Rightarrow \quad \frac{r !(5-r) !}{5 !}+\frac{r !(6-r) !}{6 !}=\frac{r !(4-r) !}{4 !}$
$\Rightarrow \quad \frac{(5-r) !}{5}+\frac{(6-r) !}{30}=(4-r) !$
$\Rightarrow \quad \frac{(5-r)}{5}+\frac{(6-r)(5-r)}{30}=1$
$\Rightarrow \quad 30-6 r+30-11r+r^{2}=30$
$\Rightarrow \quad r^{2}-17 r+30=0$
$\Rightarrow \quad r^{2}-15 r-2 r+30=0$
$\Rightarrow \quad r(r-15)-2(r-15)=0$
$\Rightarrow \quad(r-2)(r-15)=0$
$\therefore \quad r=15,2$
Hence, the value of $r$ is 2 .
$\Rightarrow \quad \frac{r !(5-r) !}{5 !}+\frac{r !(6-r) !}{6 !}=\frac{r !(4-r) !}{4 !}$
$\Rightarrow \quad \frac{(5-r) !}{5}+\frac{(6-r) !}{30}=(4-r) !$
$\Rightarrow \quad \frac{(5-r)}{5}+\frac{(6-r)(5-r)}{30}=1$
$\Rightarrow \quad 30-6 r+30-11r+r^{2}=30$
$\Rightarrow \quad r^{2}-17 r+30=0$
$\Rightarrow \quad r^{2}-15 r-2 r+30=0$
$\Rightarrow \quad r(r-15)-2(r-15)=0$
$\Rightarrow \quad(r-2)(r-15)=0$
$\therefore \quad r=15,2$
Hence, the value of $r$ is 2 .
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