Question

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

Answer 1

$3 n $ type $ \rightarrow 3,6,9= P $
$3 n -1 $ type $ \rightarrow 2,5= Q$
$3 n -2 $ type $ \rightarrow 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 $3=4+14+22+22+14+4=80$.