The Inclusion-Exclusion Principle — Three Sets

The same idea extends to three overlapping sets $A$, $B$, $C$, though the bookkeeping is a little more involved: simply adding all three individual counts and subtracting all three pairwise overlaps ends up subtracting the "all three at once" region too many times, so it has to be added back in once more.

To see why the final $+\,n(A\cap B\cap C)$ term is needed: an element belonging to all three sets is counted once in each of $n(A)$, $n(B)$, $n(C)$ (three times total), then subtracted once in each of the three pairwise-overlap terms (three times total) — leaving it counted $3 - 3 = 0$ times, when it should be counted exactly once. Adding $n(A\cap B\cap C)$ back in restores that missing count. …