Question

If $\frac{1}{{}^4{C}_n}=\frac{1}{{}^5{C}_n}+\frac{1}{{}^6{C}_n}$, then $n=$

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is B $2$

$\frac{1}{4_{C_n}}=\frac{1}{5_{C_n}}+\frac{1}{6_{C_n}}$

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

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

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

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

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

$\Rightarrow (5-n)\left[\frac{12-n}{30}\right]=1$

$\Rightarrow (5-n)(12-n)=30$

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

$\Rightarrow n=15$ or $n=2$ $\because$ $n$ cannot be greater than $4$

$\therefore$ $n=2$ [B]