Question

If$a,b$ and $c$ are the greatest values of ${}^{19}C_p,{}^{20}C_q,{}^{21}C_r$, then:

• A. $(a/11)=(b/22)=(c/42)$
• B. $(a/10)=(b/11)=(c/42)$
• C. $(a/11)=(b/22)=(c/21)$
• D. $(a/10)=(b/11)=(c/21)$

Answer 1

The correct option is A

$(a/11)=(b/22)=(c/42)$

Explanation of the correct option :

Step 1: Expressing the given data:

Given that $a,b$ and $c$ are the greatest values of ${}^{19}C_p,{}^{20}C_q,{}^{21}C_r$.

We know that

${}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}$is maximum when $\mathit{r}\mathbf{=}\{\begin{array}{ll}\frac{n}{2}& ,niseven\\ \frac{n-1}{2}or\frac{n+1}{2}& ,nisodd\end{array}$

Therefore, max $\left({}^{19}C_p\right)=\left({}^{19}C_9\right)=a$

max $\left({}^{20}C_q\right)={}^{20}C_{10}=b$

max $\left({}^{21}C_r\right)={}^{21}C_{11}=c$

Step 2: Find the relation :

$$\begin{gathered} \therefore \frac{a}{{}^{19}C_9}=\frac{b}{{\displaystyle \frac{20}{10}}\times {}^{19}C_9}=\frac{c}{{\displaystyle \frac{21}{11}}\times \frac{20}{10}\times {}^{19}C_9} \\ \Rightarrow \frac{a}{1}=\frac{b}{2}=\frac{c}{{\displaystyle \frac{42}{11}}} \\ \Rightarrow \frac{a}{11}=\frac{b}{22}=\frac{c}{{\displaystyle 42}} \end{gathered}$$

Hence, the correct option is A.