Let S = 1, 2, 3, 4, 5, 6, 9 . Then the number of elements in the...

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

Let ${N}_{p} =$Number of subsets of $S$ containing $p$ elements which are not divisible by$= 3$

For $p = 1$
$n ( {N}_{1} ) = C1 2 + C1 2 = 4$

For $p = 2$

$n ( {N}_{2} ) = C1 3 C1 2 + C1 3 C1 2 + C2 2 + C2 2 = 14$

For $p = 3$
$n ( {N}_{3} ) = C1 3 \left(C2 2 + C2 2\right) + C2 3 \left(C1 2 + C1 2\right) + C2 2 C1 2 + C1 2 C2 2 = 22$

For $p = 4$
$n ( {N}_{4} ) = C1 3 \left[\left(C2 2 C1 2\right) + \left(C1 2 C2 2\right)\right] + C2 3 \left(C2 2 + C2 2\right) + C3 3 \left(C1 2 + C1 2\right) = 22$

For $p = 5$
$n ( {N}_{5} ) = C2 3 \left[\left(C2 2 C1 2\right) + \left(C2 2 {C 2}_{2}\right)\right] + C3 3 \left(C2 2 + C2 2\right) = 14$

For $p = 6$
$n ( {N}_{6} ) = \leftC3 3 \left[( C2 2 C1 2\right) + \left(C1 2 C2 2\right)] = 4$

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