Question

Prove n(AUB) =n(A) +n(B)
-n(A intersection B)​

Answer 1

Answer:

Step-by-step explanation:

➡️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

n(A U B) = n(A) + n(B) - n(A ∩ B)

Hence proved ..

Hope it helps..

Brainliest please..

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{hello}}}}}}$

➡️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

n(A U B) = n(A) + n(B) - n(A ∩ B)