Question

If ${}^{19}{C}_r$ and ${}^{19}{C}_{r-1}$ are in the ratio 2:3, then find ${}^{14}{C}_r$

• A. 91
• B. 81
• C. 71
• D. 61

Answer 1

The correct option is A 91

We know that, ${n_C}_r=\frac{n!}{r!(n-r)!}$
Given, $\frac{{{19}_C}_r}{{{19}_C}_{r-1}}=\frac{2}{3}$
$=>\frac{\frac{19!}{r!(19-r)!}}{\frac{19!}{(r-1)!(19-r+1)!}}=\frac{2}{3}$
$=>\frac{(r-1)!(20-r)!}{r!(19-r)!}=\frac{2}{3}$
$=\frac{(r-1)!\times (20-r)\times (19-r)!}{r\times (r-1)!\times (19-r)!}=\frac{2}{3}$
$=>\frac{20-r}{r}=\frac{2}{3}$
$=>r=12$
So,
${{14}_C}_{12}=\frac{14!}{12!(14-12)!}=\frac{14!}{12!\left(2\right)!}=\frac{14\times 13\times 12!}{12!\times 2}=7\times 13=91$