JEE Main 2021 — Mathematics Algebra

We can categorise the numbers of set $S={\underset{3n\text{ type}}{\underset{⏟}{3,6,9}},\underset{3(n-1)\text{ type}}{\underset{⏟}{2,5}},\underset{3(n-2)\text{ type}}{\underset{⏟}{1,4}}}$S=3n type⏟3,6,9​​,3(n−1) type⏟2,5​​,3(n−2) type⏟1,4​​Let Np=Number of subsets of $S$S containing $p$p elements which are not divisible by$=3$=3

For $p=1$p=1

$$n({N}_{1})=C12+C12=4$$n(N1​)=C12+C12=4

For $p=2$p=2

$$n({N}_{2})=C13C12+C13C12+C22+C22=14$$n(N2​)=C13C12+C13C12+C22+C22=14

For $p=3$p=3

$$n({N}_{3})=C13(C22+C22)+C23(C12+C12)+C22C12+C12C22=22$$n(N3​)=C13(C22+C22)+C23(C12+C12)+C22C12+C12C22=22

For $p=4$p=4

$$n({N}_{4})=C13[(C22C12)+(C12C22)]+C23(C22+C22)+C33(C12+C12)=22$$n(N4​)=C13[(C22C12)+(C12C22)]+C23(C22+C22)+C33(C12+C12)=22

For $p=5$p=5

$$n({N}_{5})=C23[(C22C12)+(C22{C2}_{2})]+C33(C22+C22)=14$$n(N5​)=C23[(C22C12)+(C22C22​)]+C33(C22+C22)=14

For $p=6$p=6

$$n({N}_{6})=C33[(C22C12)+(C12C22)]=4$$n(N6​)=C33[(C22C12)+(C12C22)]=4

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