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

3n type $\to$ 3, 6, 9 = P<br><br>3n $-$ 1 type $\to$ 2, 5 = Q<br><br>3n $-$ 2 type $\to$ 1, 4 = R<br><br>number of subset of S containing one element which are not divisible by 3 = ${}^2$C₁ + ${}^2$C₁ = 4<br><br>number of subset of S containing two numbers whose some is not divisible by 3<br><br>= ${}^3$C₁ $\times$ ${}^2$C₁ + ${}^3$C₁ $\times$ ${}^2$C₁ + ${}^2$C₂ + ${}^2$C₂ = 14<br><br>number of subsets containing 3 elements whose sum is not divisible by 3<br><br>= ${}^3$C₂ $\times$ ${}^4$C₁ + (${}^2$C₂ $\times$ ${}^2$C₁)2 + ${}^3$C₁(${}^2$C₂ + ${}^2$C₂) = 22<br><br>number of subsets containing 4 elements whose sum is not divisible by 3<br><br>= ${}^3$C₃ $\times$ ${}^4$C₁ + ${}^3$C₂(${}^2$C₂ + ${}^2$C₂) + (${}^3$C₁${}^2$C₁ $\times$ ${}^2$C₂)2<br><br>= 4 + 6 + 12 = 22<br><br>number of subsets of S containing 5 elements whose sum is not divisible by 3.<br><br>= ${}^3$C₃(${}^2$C₂ + ${}^2$C₂) + (${}^3$C₂${}^2$C₁ $\times$ ${}^2$C₂) $\times$ 2 = 2 + 12 = 14<br><br>number of subsets of S containing 6 elements whose sum is not divisible by 3 = 4<br><br>$\Rightarrow$ Total subsets of Set A whose sum of digits is not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80.