Question

$A,B,C$ are three sets such that $n\left(A\right)=2$,$n\left(B\right)=3$,$n\left(C\right)=4$, If $P\left(x\right)$ denotes power set of $X$,$k=\frac{n\left(P\left(P\left(C\right)\right)\right)}{n\left(P\left(P\left(A\right)\right)\right)\times n\left(P\left(P\left(B\right)\right)\right)}$.then find the Sum of digits of $k$

Answer 1

$n\left(A\right)=2\Rightarrow P\left(A\right)=2^2=4\Rightarrow P\left(\left(A\right)\right)=2^4$

$n\left(B\right)=3\Rightarrow P\left(B\right)=2^3=8\Rightarrow P\left(\left(B\right)\right)=2^8$

$n\left(C\right)=4\Rightarrow P\left(C\right)=2^4=16\Rightarrow P\left(\left(A\right)\right)=2^{16}$

$K=\frac{n\left(P\left(P\left(C\right)\right)\right)}{n\left(P\left(P\left(A\right)\right)\right)\times n\left(P\left(P\left(B\right)\right)\right)}$

$=\frac{2^{16}}{2^4\times 2^8}=2^{16-12}=2^4=16$

Sum of the digits$=1+6=7$