Question

If ${}^n{C}_{r-1}:{}^n{C}_r:{}^n{C}_{r+1}=2:3:4$ then $(n,r)$

• A. $(11,7)$
• B. $(15,10)$
• C. $(34,14)$
• D. $(24,16)$

Answer 1

Answer: (C)

$(34,14)$

$\frac{{{}^{n}C}_{r-1}}{{{}^{n}C}_r}=\frac{2}{3}$

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

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

$\frac{r}{n-r+1}=\frac{2}{3}$

$3r=2n-2r+2$

$5r=2n+2⟶\left(1\right)$

$\Rightarrow \frac{{{}^{n}C}_r}{{{}^{n}C}_{r+1}}=\frac{3}{4}$

$\frac{\frac{n!}{(n-r)!r!}}{\frac{n!}{(n-r-1)(r+1)!}}=\frac{3}{4}$

$\frac{(n-r-1)!}{(n-r)!}\frac{(r+1)!}{r!}=\frac{3}{4}$

$\frac{r+1}{n-r}=\frac{3}{4}$

$4r+4=3n-3r$

$7r=3n-4⟶\left(2\right)$

Solving $⟶\left(1\right)$ $\&$ $⟶\left(2\right)$

$\Rightarrow 5r=2n+2\times 3$

$7r=3n-4\times 2$

$\Rightarrow 5\times 14=2n+2$

$\Rightarrow n=\frac{68}{2}$

$=34$

$\Rightarrow (n,r)=(34,14).$

Hence, the answer is $(n,r)=(34,14).$