Question

A={1,3,5},B={2,3,5,6} and C= {1,5,6,7}.verify n (AUBUC)= n(A)+nAU n(B) +n(C) -n(AnB)- n (BnC)-n (AnC)+n (AnBnC)​

Answer 1

Chapter - Sets

Given: A = {1, 3, 5}, B = {2, 3, 5, 6}, C = {1, 5, 6, 7}

To prove: n(A U B U C) = n(A) + n(B) + n(C) - n(A∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)

Proof:

• A U B U C = {1, 2, 3, 5, 6, 7}
• Then, n(A U B U C) = 6

• A = {1, 3, 5}
• Then, n(A) = 3

• B = {2, 3, 5, 6}
• Then, n(B) = 4

• C = {1, 5, 6, 7}
• Then, n(C) = 4

• A ∩ B = {3, 5}
• Then, n(A ∩ B) = 2

• B ∩ C = {5, 6}
• Then, n(B ∩ C) = 2

• C ∩ A = {1, 5}
• Then, n(C ∩ A) = 2

• A ∩ B ∩ C = {5}
• Then, n(A ∩ B ∩ C) = 1

Now, R.H.S. = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

= 3 + 4 + 4 - 2 - 2 - 2 + 1

= 6 = n(A U B U C) = L. H. S.

Hence proved.

This completes the verification.