Question

n(AuB)=n(A)+n(B)=n(A∩B)
Solve it please

Answer 1

Answer:

n(AuB)=n(A) + n(B) - n(A intersection B)

therefore

n(AuB)= n(A) +n(B)=n(A intersection B)

Step-by-step explanation:

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Answer 2

I thing there is some mistake in question.

the question should be...

n(AuB)=n(A)+n(B)-n(A intesection B)

then the answer is come....

A(1,2,3,4) B(2,4,5,6)

then n(AuB) = (1,2,3,4,5,6)

n(A)=(1,2,3,4) + n(B)=(2,4,5,6) then it become n(A+B) = (1,2,2,3,4,4,5,6)

then n(A+B)-n(A intersection B) = (1,2,2,3,4,4,5,6) - (2,4)

then its is equal  to LHS = (1,2,3,4,5,6)