bhushangiri wrote:

It should be A + B + C - AB - BC - CA + ABC -- when AB BC and CA include intersect of all 3. Note that its only ABC and not 3ABC as mentioned in the previous post.

The formula you quoted is correct if you are considering AB = "A and B alone" no C involved. And similar logic for BC and CA

how can I interpret this logically?

We remove AB because we double count them in A, and B.

We add back ABC because we "double remove" them in AB, BC, CA.

Is that sort of the logic?

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