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 $9$

${}^n{C}_{r-1}=\frac{n!}{(r-1)!(n-r+1)!}..\left(1\right)$

${}^n{C}_r=\frac{n!}{r!(n-r)!}..\left(2\right)$

${}^n{C}_{r+1}=\frac{n!}{(r+1)!(n-r-1)!}..\left(3\right)$

Dividing equation $\left(2\right)$ by $\left(1\right)$, we get

$\frac{84}{36}=\frac{n-r+1}{r}$

$\therefore 7r=3n-3r+3$

$\therefore 10r=3n+3..\left(4\right)$

Dividing equation $\left(3\right)$ by $\left(2\right)$, we get

$\frac{126}{84}=\frac{n-r}{r+1}$

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

$\therefore 2n=5r+3..\left(5\right)$

Multiplying equation $\left(5\right)$ by $2$ and adding that to equation $\left(4\right)$, we get

$4n+10r=3n+3+10r+6$

$\therefore n=9$

$\therefore {}^n{C}_8={}^9{C}_8=9$