Question

If ${}^n{C}_{r-1}=36$ and ${}^n{C}_r=84$, then

• A. $13r-3n-3=0$
• B. $10r-3n-30=0$
• C. $10r+3n-3=0$
• D. $10r-3n+3=0$
• E. $10r-3n-3=0$

Answer 1

Answer: (E)

Given, ${}^n{C}_{r-1}=36$
and ${}^n{C}_r=84$
Now, $\frac{{}^n{C}_r-1}{{}^n{C}_r}=\frac{36}{84}$
$\Rightarrow \frac{r}{n-r+1}=\frac{36}{84}=\frac{3}{7}$
$\Rightarrow 7r=3n-3r+3$
$\Rightarrow 10r=3n+3$
$\Rightarrow 10r-3n-3=0$