Question

Let $S=\{1,2,3,4,5,6,9\}.$ Then the number of elements in the set $T=\{A\subseteq S:A\ne \varphi$ and the sum of all the elements of $A$ is not a multiple of 3$\}$ is _____

Answer 1

Answer: 80

$3n$ type $\to 3,6,9=P$
$3n-1$ type $\to 2,5=Q$
$3n-2$ type $\to 1,4=R$
number of subset of $S$ containing one element
which are not divisible by $3={}^2{C}_1+{}^2{C}_1=4$
number of subset of S containing two numbers whose some is not divisible by 3
$={}^3{C}_1\times {}^2{C}_1+{}^3{C}_1\times {}^2{C}_1+{}^2{C}_2+{}^2{C}_2=14$
number of subsets containing 3 elements whose sum is not divisible by 3
$={}^3{C}_2\times {}^4{C}_1+\left({}^2{C}_2\times {}^2{C}_1\right)2+{}^3{C}_1\left({}^2{C}_2+{}^2{C}_2\right)=22$
number of subsets containing 4 elements whose sum is not divisible by 3
$={}^3{C}_3\times {}^4{C}_1+{}^3{C}_2\left({}^2{C}_2+{}^2{C}_2\right)+\left({}^3{C}_1{}^2{C}_1\times {}^2{C}_2\right)2$
$=4+6+12=22$
number of subsets of $S$ containing 5 elements whose sum is not divisible by 3 .
$={}^3{C}_3\left({}^2{C}_2+{}^2{C}_2\right)+\left({}^3{C}_2{}^2{C}_1\times {}^2{C}_2\right)\times 2$
$=2+12=14$
number of subsets of $S$ containing 6 elements whose sum is not divisible by $3=4$
$\Rightarrow$ Total subsets of Set A whose sum of digits is not divisible by $3=4+14+22+22+14+4=80$.