Question

Let S = {1, 2, 3, 4, 5, 6, 9}. Then the number of elements in the set T = {A ⊆ S : A ≠ ϕ and the sum of all the elements of A is not a multiple of 3} is ______.

Answer 1

3n type → 3, 6, 9 = P 

3n – 1 type → 2, 5 = Q 

3n – 2 type → 1,4 = R

number of subset of S containing one element which are not divisible by 3 = ²C₁ + ²C₁ = 4 number of subset of S containing two numbers whose some is not divisible by 3 

= ³C₁ x ²C₁ + ³C₁ x ²C₁ + ²C₂ + ²C₂ = 14

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

= ³C₂ x ⁴C₁ + (²C₂ x ²C₁)2 + ³C₁(²C₂+ ²C₂) = 22

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

= ³C₂ x ⁴C₁ + ³C₂(²C₂ + ²C₂) + (³C₁²C1+ ²C₂)2

= 4 + 6 + 12 = 22. 

number of subsets of S containing 5 elements whose sum is not divisible by 3. 

= ³C₃(²C₂ + ²C₂) + (³C₂²C₁ x ²C₂) x 2

= 2 + 12 = 14 

number of subsets of S containing 6 elements whose sum is not divisible by 3 = 4 

⇒ Total subsets of Set A whose sum of digits is not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80.

Answer 2

There are two numbers of the type 3λ + 1, two numbers of the type 3λ – 1 and three numbers of the type 3λ.

So, number of subsets whose sum is divisible by 3

= 2³⋅(²C₀² + ²C₁² + ²C₂²)

= 48

Required number of subsets = 27 − 48 = 80