Question

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

• A. $10$
• B. $7$
• C. $9$
• D. $8$

Answer 1

The correct option is C $9$

As we know,
${}^n{C}_{r-1}=\frac{n!}{(r-1)!(n-r+1)!}=\mathrm{36...}\left(1\right)$
${}^n{C}_r=\frac{n!}{r!(n-r)!}=\mathrm{84...}\left(2\right)$
${}^n{C}_{r+1}=\frac{n!}{(r+1)!(n-r-1)!}=\mathrm{126...}\left(3\right)$
Divide the equation (1) by (2)
$\frac{r}{n-r+1}=\frac{36}{84}$
$\Rightarrow 3n-10r+3=\mathrm{0...}\left(4\right)$
Now, divide the equation (2) by (3)
$\frac{r+1}{n-r}=\frac{84}{126}$
$\Rightarrow 3r+3=2n-2r$
$\Rightarrow 2n-5r-3=\mathrm{0...}\left(5\right)$
On solving equation (4) and (5), get,
$n=9,r=3$
So, ${}^n{C}_8={}^9{C}_8=9$