Question

If $\frac{1}{{}^{4}C_n}=\frac{1}{{}^{5}C_n}+\frac{1}{{}^{6}C_n}$ then $n$ is equal to?

• A. $3$
• B. $2$
• C. $1$
• D. $0$
• E. $4$

Answer 1

The correct option is B

$2$

Step-1: Simplify the given data.

Given: $\frac{1}{{}^{4}C_n}=\frac{1}{{}^{5}C_n}+\frac{1}{{}^{6}C_n}$

$\Rightarrow$$\frac{1}{{\displaystyle \frac{4!}{\left(4-n\right)!n!}}}=\frac{1}{{\displaystyle \frac{5!}{\left(5-n\right)!n!}}}+\frac{1}{{\displaystyle \frac{6!}{\left(6-n\right)!n!}}}$

$\Rightarrow$ $\frac{\left(4-n\right)!n!}{{\displaystyle 4!}}=\frac{\left(5-n\right)!n!}{{\displaystyle 5!}}+\frac{\left(6-n\right)!n!}{{\displaystyle 6!}}$

$\Rightarrow$ $\frac{\left(4-n\right)!}{{\displaystyle 4!}}=\frac{\left(5-n\right)!}{{\displaystyle 5!}}+\frac{\left(6-n\right)!}{{\displaystyle 6!}}$

$\Rightarrow$ $\frac{\left(4-n\right)!}{{\displaystyle 4!}}=\frac{\left(5-n\right)\times \left(4-n\right)!}{{\displaystyle 5!}}+\frac{\left(6-n\right)\times \left(5-n\right)\times \left(4-n\right)!}{{\displaystyle 6!}}$

$\Rightarrow$ $1=\frac{\left(5-n\right)}{{\displaystyle 5}}+\frac{\left(6-n\right)\times \left(5-n\right)}{{\displaystyle 6\times 5}}$

$\Rightarrow$ $1=\left(5-n\right)\left\{\frac{6+6-n}{{\displaystyle 30}}\right\}$

$\Rightarrow$ $30=\left(5-n\right)\left\{12-n\right\}$

$\Rightarrow$ $0=n^2-17n+30$

Step-2: Solve the quadratic equation.

$\Rightarrow$ $0=n^2-15n-2n+30$

$\Rightarrow$ $0=\left(n-15\right)\left(n-2\right)$

$\Rightarrow$ $n=15,2$

Hence, correct answer is option $\left(B\right)$.