Question

If $a,b$ and $c$ are the greatest values of ${}^{19}{C}_p,{}^{20}{C}_q,{}^{21}{C}_r$ respectively, then:

• A. $\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$
• B. $\frac{a}{10}=\frac{b}{11}=\frac{c}{42}$
• C. $\frac{a}{11}=\frac{b}{22}=\frac{c}{21}$
• D. $\frac{a}{10}=\frac{b}{11}=\frac{c}{21}$

Answer 1

The correct option is A $\frac{a}{11}=\frac{b}{22}=\frac{c}{42}$

We know that, ${}^n{C}_r$ is maximum when $r=\{\begin{array}{c}\frac{n}{2},\text{n is even}\\ \frac{n+1}{2}\text{or}\frac{n-1}{2},\text{n is odd}\end{array}$

Therefore, $max{(}^{19}{C}_p)={}^{19}{C}_9=a$
$max{(}^{20}{C}_q)={}^{20}{C}_{10}=b$
$max{(}^{21}{C}_r)={}^{21}{C}_{11}=c$

$\therefore \frac{a}{{}^{19}{C}_9}=\frac{b}{{}^{20}{C}_{10}}=\frac{c}{{}^{21}{C}_{11}}=1$
$\Rightarrow \frac{a}{{}^{19}{C}_9}=\frac{b}{\frac{20}{10}\times {}^{19}{C}_9}=\frac{c}{\frac{21}{11}\times \frac{20}{10}\times {}^{19}{C}_9}=1$
$\Rightarrow \frac{a}{1}=\frac{b}{2}=\frac{c}{\frac{42}{11}}$
$\Rightarrow \frac{a}{11}=\frac{b}{22}=\frac{c}{42}$