Cardinality and the Inclusion-Exclusion Principle

The cardinality of a finite set $A$, written $n(A)$, is simply the count of its distinct elements. $n(\emptyset) = 0$; if $A$ is a singleton, $n(A) = 1$.

A natural question follows: given $n(A)$ and $n(B)$ separately, can we find $n(A \cup B)$? Simply adding $n(A) + n(B)$ over-counts every element lying in both sets, since it gets counted once inside $n(A)$ and again inside $n(B)$. Correcting for that double count gives the inclusion-exclusion principle for two sets:

The same idea extends to three sets, where elements can now be over-counted in more than one way — every pairwise overlap is subtracted once, but the triple overlap, having been subtracted three times over (once within each pairwise term), must be added back exactly once:

Worked illustration. Suppose a market survey of 120 customers of an Odisha handicrafts store finds 70 customers buy textile products ($A$), 50 buy pottery products ($B$), and 25 buy both. Then

customers buy at least one of the two, and $120 - 95 = 25$ customers buy neither.

<!-- FIGURE-NEEDED: Venn diagram, rectangle U labelled 120 total, two overlapping circles A (textile, only-region shows 45) and B (pottery, only-region shows 25), overlap region shows 25, and 25 shown outside both circles but inside U, illustrating the worked inclusion-exclusion example --> …