Prove that for any two set A and B, n(AuB)=n(A)+n(B)-n(AnB)

1.	Prove that for any two set A and B, n(AuB)=n(A)+n(B)-n(AnB)
Answer» > Here from vendiagram n(a-B) =n(a)-n(anb) And, n(b-a) =n(b)-n(anb) Again from vendiagram we get 1 formula n(aub)=n(a-b)+n(b-a)+n(a-b) n(aub)=n(a)+n(b)-n(anb) [putting the VALUE of n(a-b), n(b-a) ] Hence proved. I hope my answer is HELPFUL for you ✌️☺️

Answer»

Here from vendiagram

n(a-B) =n(a)-n(anb)

And, n(b-a) =n(b)-n(anb)

Again from vendiagram we get 1 formula

n(aub)=n(a-b)+n(b-a)+n(a-b)

n(aub)=n(a)+n(b)-n(anb) [putting the VALUE of n(a-b), n(b-a) ]

Hence proved.

I hope my answer is HELPFUL for you ✌️☺️

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Prove that for any two set A and B, n(AuB)=n(A)+n(B)-n(AnB)​

Here from vendiagram

n(a-B) =n(a)-n(anb)

And, n(b-a) =n(b)-n(anb)

Again from vendiagram we get 1 formula

n(aub)=n(a-b)+n(b-a)+n(a-b)

n(aub)=n(a)+n(b)-n(anb) [putting the VALUE of n(a-b), n(b-a) ]

Hence proved.

I hope my answer is HELPFUL for you ✌️☺️

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Prove that for any two set A and B, n(AuB)=n(A)+n(B)-n(AnB)