Question

Which of the following option is the correct combination?

• A. $I,iv,S$
• B. $III,ii,R$
• C. $IV,I,P$
• D. $III,iii,Q$

Answer 1

The correct option is C $IV,I,P$

${}^n{C}_r=\frac{n!}{(n-r)!r!}$, here $S=19,R=18,P=16$

$fx\left(A\right),{}^1{C}_4=\frac{1!}{(1-4)!4!}$ not possible

$fx\left(B\right),{}^3{C}_2=\frac{3!}{2!1!}=3=18$ this is possible only when $3\left(6\right)=18$

but $6^3{C}_2$ i.e. $2n^n{C}_r$, also this does not given all $\left(4\right)$ option

SO, $n^n{C}_r$ proves the option $\left(C\right)$;

i.e., $4^4{C}_1=4\frac{4!}{3!!}\Rightarrow 4\left(4\right)=16=P$

$\Rightarrow$ option $\left(C\right)$ is correct