Question

prove that n(AuB)=n(A)+n(B)-n(AnB)

Answer 1

Hello!
 
→ n(A U B) = n(A only) + n(B only) + n(A ∩ B) 

→ n(A U B) = n(A) - x + n(B) - x + x 

→ n(A U B) = n(A) + n(B) - x - x + x 

→ n(A U B) = n(A) + n(B) - 2x + x 

→ n(A U B) = n(A) + n(B) - x

$\boxed{\sf n(A U B) = \sf n(A) + n(B) - n(A ∩ B)}$

Cheers!

Answer 2

Answer:

yes, above answer is correct.