Mathematics - Permutation & Combination Question with Solution | TestHub

We can categorise the numbers of set $S=\{\underset{ 3n\text{ type}}{\underbrace{3,6,9}}, \underset{ 3(n-1)\text{ type}}{\underbrace{2,5}} ,\underset{ 3(n-2)\text{ type}}{\underbrace{1,4}}\}$$$

Let $N_p=$Number of subsets of $S$ containing $p$ elements which are not divisible by$=3$

For $p=1$
$n\left(N_1\right)={}^{2}C_1+{}^{2}C_1=4$

For $p=2$

$n\left(N_2\right)={}^3\mathrm{C}_1{}^2\mathrm{C}_1+{}^3\mathrm{C}_1{}^2\mathrm{C}_1+{}^2\mathrm{C}_2+{}^2\mathrm{C}_2=14$

For $p=3$
$n\left(N_3\right)={}^3\mathrm{C}_1\left({}^2\mathrm{C}_2+{}^2\mathrm{C}_2\right)+{}^3\mathrm{C}_2\left({}^2\mathrm{C}_1+{}^2\mathrm{C}_1\right)+{}^2\mathrm{C}_2{}^2\mathrm{C}_1+{}^2\mathrm{C}_1{}^2\mathrm{C}_2=22$

For $p=4$
$n\left(N_4\right)={}^3\mathrm{C}_1\left[\left({}^2\mathrm{C}_2{}^2\mathrm{C}_1\right)+\left({}^2\mathrm{C}_1{}^2\mathrm{C}_2\right)\right]+{}^3\mathrm{C}_2\left({}^2\mathrm{C}_2+{}^2\mathrm{C}_2\right)+{}^3\mathrm{C}_3\left({}^2\mathrm{C}_1+{}^2\mathrm{C}_1\right)=22$

For $p=5$
$n\left(N_5\right)={}^3\mathrm{C}_2\left[\left({}^2\mathrm{C}_2{}^2\mathrm{C}_1\right)+\left({}^2\mathrm{C}_2{{}^2\mathrm{C}}_2\right)\right]+{}^3\mathrm{C}_3\left({}^2\mathrm{C}_2+{}^2\mathrm{C}_2\right)=14$

For $p=6$
$n\left(N_6\right)=\left.{}^3\mathrm{C}_3\left[({}^2\mathrm{C}_2{}^2\mathrm{C}_1\right)+\left({}^2\mathrm{C}_1{}^2\mathrm{C}_2\right)\right]=4$

$\Rightarrow$Total subsets satisfying given condition $=4+14+22+22+14+4=80$