Q56 / 155 · WBJEE 2016 · Mathematics

If $\frac{1}{{ }^{5} C_{r}}+\frac{1}{{ }^{6} C_{r}}=\frac{1}{{ }^{4} C_{r}},$ then the value of $r$ is

  • A
    4
  • Correct answer
    B
    2
  • C
    5
  • D
    3

Solution

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 .

Hence, the correct answer is option: 'B'.

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