Question

Observe the figure and verify the following equation:
n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)

Answer 1

I couldn't do it either.

Answer 2

we have to draw the figure and verify the following equation :

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

solution : see figure, it is the graph of given equation.

verification : n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

LHS = n(A U B UC) = n{(A U B) U C }

we know, n(x U y) = n(x) + n(y) - n(x ∩ y)

so, n{(A U B) U C} = n(A U B) + n(C) - n{(A U B) ∩ C}

now, n(A U B) = n(A) + n(B) - n(A ∩ B)

and (A U B) ∩ C = (A ∩ C) U (B ∩ C) [ from property of sets ]

now, n{(A U B) U C} = n(A) + n(B) - n(A ∩ B) + n(C) - n{(A ∩ C) U (B ∩ C)}

= n(A) + n(B) + n(C) - n(A ∩ B) - [n(A ∩ C) + n(B ∩ C) - n(A ∩ C) ∩ (B ∩ C)]

it is obvious that (A ∩ C) ∩ (B ∩ C) = A ∩ B ∩ C

= n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)

Therefore n(A U B U C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)

hence verified