JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Permutations and Combinations

Classify $S=\{1,2,3,4,5,6,9\}$ by residue mod $3$: residue $0$ is $\{3,6,9\}$, residue $1$ is $\{1,4\}$, residue $2$ is $\{2,5\}$. The three residue-$0$ elements may be chosen freely ($2^3=8$ ways). Among subsets of $\{1,4\}$ the residue counts are $1,2,1$ for sums $\equiv 0,1,2$, and among subsets of $\{2,5\}$ they are $1,1,2$. Pairing to make the total $\equiv 0$ gives $1\cdot 1+2\cdot 2+1\cdot 1=6$ ways. So $8\times 6=48$ subsets (including the empty set) have sum divisible by $3$. Of all $2^7=128$ subsets, $128-48=80$ have sum not divisible by $3$, and all of these are non-empty. The number of such subsets is $80$.