QJEEM21Mad5e6: JEE Main 2021 Mathematics - Class 11 JEEM21Mad5e6 Solution: Let S = {1, 2, 3, 4, 5, 6, 9}. Then the number of elements in…

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 = ${}^2$ C₁ + ${}^2$ C₁ = 4

number of subset of S containing two numbers whose some is not divisible by 3

= ${}^3$ C₁ $\times$ ${}^2$ C₁ + ${}^3$ C₁ $\times$ ${}^2$ C₁ + ${}^2$ C₂ + ${}^2$ C₂ = 14

number of subsets containing 3 elements whose sum is not divisible by 3

= ${}^3$ C₂ $\times$ ${}^4$ C₁ + ( ${}^2$ C₂ $\times$ ${}^2$ C₁)2 + ${}^3$ C₁( ${}^2$ C₂ + ${}^2$ C₂) = 22

number of subsets containing 4 elements whose sum is not divisible by 3

= ${}^3$ C₃ $\times$ ${}^4$ C₁ + ${}^3$ C₂( ${}^2$ C₂ + ${}^2$ C₂) + ( ${}^3$ C₁ ${}^2$ C₁ $\times$ ${}^2$ C₂)2

= 4 + 6 + 12 = 22

number of subsets of S containing 5 elements whose sum is not divisible by 3.

= ${}^3$ C₃( ${}^2$ C₂ + ${}^2$ C₂) + ( ${}^3$ C₂ ${}^2$ C₁ $\times$ ${}^2$ C₂) $\times$ 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.