Question

Given that C(n, r) : C ( n, r + 1)=1 : 2 and C (n, r + 1) : C (n, r + 2) = 2 : 3.

Find $n.$

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

The correct option is D $14$

Given : $\frac{{}_r^{n}C}{{}_{r+1}^{n}C}=\frac{1}{2}$

$\Rightarrow 2\frac{n!}{r!(n-r)!}=\frac{n!}{(r+1)!(n-r-1)!}$

$\Rightarrow 3r+2=n$

Also, $\frac{{}_{r+1}^{n}C}{{}_{r+2}^{n}C}=\frac{2}{3}$

$\Rightarrow 3\frac{n!}{(r+1)!(n-r-1)!}=2\frac{n!}{(r+2)!(n-r-2)!}$

$\Rightarrow 3r+6=2n-2r-2$

$\Rightarrow 5r+8=2n$

Substituting $n$, we get

$8+5r=2(3r+2)$

$8+5r=6r+4$

$\Rightarrow r=4$

Solving for $n$, we get

$n=3\left(4\right)+2=12+2$

$n=14$

Hence, D is correct.