Correct Answer – Option 2 : 1 and 3 only

Concept:

Let U be the universal set and A, B, C be the subsets of U.

If \(\rm A∩ B ∩ C = \phi\) then  A, B, C are mutually exclusive.

Çalculations:

Given, U = {1, 2, 3, …., 20}.

Let A, B, C be the subsets of U.

A be the set of all numbers which are perfect squares

⇒ A = {1, 4, 9 16}

B be the set of all numbers which are multiples of 5

⇒ B = {5, 10, 15, 20}

and C be the set of all numbers, which are divisible by 2 and 3

⇒ C = {6, 12, 18}

Now, \(\rm A∩ B ∩ C = \phi\)

So, Å, B, C are mutually exclusive.

Hence statement 1 is correct

Here  Å, B, C are mutually exclusive so A, B, C can’t be mutually exhaustive

Hence statement 2 is wrong

A ∪ B = {1, 4, 5, 9, 10, 15, 16, 20}

n(A ∪ B) = 8

U = {1, 2, 3, …., 20}

n(U) = 20

Now, The number of elements in the complement set of A ∪ B  = n(U) – n(A ∪ B) = 20 – 8 = 12

Hence statement 3 is correct