# Solve the Following Questions

Question:

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

Solution:

$3 \mathrm{n}$ type $\rightarrow 3,6,9=\mathrm{P}$

$3 \mathrm{n}-1$ type $\rightarrow 2,5=\mathrm{Q}$

$3 \mathrm{n}-2$ type $\rightarrow 1,4=\mathrm{R}$

number of subset of $\mathrm{S}$ containing one element which are not divisible by $3={ }^{2} \mathrm{C}_{1}+{ }^{2} \mathrm{C}_{1}=4$ number of subset of $S$ containing two numbers whose some is not divisible by 3

$={ }^{3} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{1}+{ }^{3} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{1}+{ }^{2} \mathrm{C}_{2}+{ }^{2} \mathrm{C}_{2}=14$

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

$={ }^{3} \mathrm{C}_{2} \times{ }^{4} \mathrm{C}_{1}+\left({ }^{2} \mathrm{C}_{2} \times{ }^{2} \mathrm{C}_{1}\right) 2+{ }^{3} \mathrm{C}_{1}\left({ }^{2} \mathrm{C}_{2}+{ }^{2} \mathrm{C}_{2}\right)=22$

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

$={ }^{3} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{1}+{ }^{3} \mathrm{C}_{2}\left({ }^{2} \mathrm{C}_{2}+{ }^{2} \mathrm{C}_{2}\right)+\left({ }^{3} \mathrm{C}_{1}{ }^{2} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{2}\right) 2$

$=4+6+12=22$

number of subsets of $\mathrm{S}$ containing 5 elements whose sum is not divisible by 3 .

$={ }^{3} \mathrm{C}_{3}\left({ }^{2} \mathrm{C}_{2}+{ }^{2} \mathrm{C}_{2}\right)+\left({ }^{3} \mathrm{C}_{2}{ }^{2} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{2}\right) \times 2=2+12=14$

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 $3=4+14+22+22+14+4=80$.