Question

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

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

The correct option is B 2

$\frac{1}{{}^4{C}_n}=\frac{1}{{}^5{C}_n}+\frac{1}{{}^6{C}_n}$

$\frac{1}{\frac{4!}{n!(4-n)!}}=\frac{1}{\frac{5!}{n!(5-n)!}}+\frac{1}{\frac{6!}{n!(6-n)!}}$ $[{\because }^n{C}_r=\frac{n!}{r!(n-r)!}]$

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

$\frac{n!(4-n)!}{4!}=\frac{n!(5-n)(5-n-1)!}{5\times 4!}+\frac{n!(6-n)(6-n-1)(6-n-2)!}{6\times 5\times 4!}$

$\Rightarrow \frac{n!(4-n)!}{4!}=\frac{n!(4-n)!(5-n)}{5.4!}+\frac{n!(4-n)!(6-n)(5-n)}{30.4!}$

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

$\Rightarrow 30=6(5-n)+(6-n)(5-n)$

$\Rightarrow 30=30-6n+30-6n-5n+n^2$

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

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

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

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

$\therefore n=2$ or $n=15$, which is rejected since $n>4$

$\therefore n=2$.