Question

If ${}^n{C}_{r-1}=36{,}^n{C}_r=84$ and ${}^n{C}_{r+1}=126$ then the value of ${}^n{C}_8$

• A. $10$
• B. $7$
• C. $9$
• D. $8$

Answer 1

Answer: (C)

$\frac{n}{r-1n-r+1}=\mathrm{36.....}\left(1\right)$
$\frac{n}{rn-r}=\mathrm{84...........}\left(2\right)$
$\frac{n}{r+1n-r-1}=\mathrm{126......}\left(3\right)$
$\left(1\right)÷\left(2\right)$ gives $\frac{r}{n-r+1}=\frac{36}{84}\Rightarrow 84r=36n-36r+36$ or $120r=36n+\mathrm{36......}\left(4\right)$
$\left(2\right)÷\left(3\right)$ gives $\frac{r+1}{n-r}=\frac{84}{126}\Rightarrow 126r+126=84n-84r$ or$210r=84n-\mathrm{126...........}\left(5\right)$
Solving $\left(4\right)$ and $\left(5\right)n=9,r=3$
So ${}^n{C}_8{=}^9{C}_8=9$