The systematic way to count a union when the sets overlap.
The formula
For two sets: |A or B| = |A| + |B| - |A and B|.
For three: |A| + |B| + |C| - |AB| - |AC| - |BC| + |ABC|.
In general, alternate: add all singles, subtract all pairs, add all triples, and so on.
Try the complement first
Before reaching for the full machinery, check whether the complement is easier. Questions phrased "at least one" almost always are:
P(at least one) = 1 - P(none)
Probability that at least one of four dice shows a six? Do not enumerate. 1 - (5/6)^4 ≈ 0.518.
That single move handles a large fraction of interview questions that look like inclusion-exclusion problems.
Derangements
The canonical application. A derangement is a permutation with no fixed point - nobody gets their own hat back.
Applying inclusion-exclusion over the events "person k gets their own hat":
D(n)/n! = 1 - 1/1! + 1/2! - 1/3! + ... + (-1)^n / n!
That series is the expansion of e^(-1), so:
P(no one gets their own hat) approaches 1/e ≈ 0.368
This converges startlingly fast - it is already accurate to three decimals at n = 6. Contrast it with the expected number of matches, which is exactly 1 for every n. Both facts about the same problem, and interviewers like asking for them together.
The matching-birthday cousin
Inclusion-exclusion also gives exact answers for problems the birthday problem approximates, though the complement route is far quicker there.
When to stop
The number of terms doubles with each set. With four or more overlapping conditions the formula is technically correct and practically hopeless in an interview. If you find yourself writing the fourth level of terms, there is almost certainly a symmetry or a complement you have missed.
More in combinatorics.