C40 — count only integers divisible by exactly one of 2 or 5
D55 — subtract 15 for the overlap, since multiples of 10 appear in both
The intersection A ∩ B is the set of integers divisible by both 2 and 5, i.e., divisible by lcm(2,5) = 10. There are 10 such integers (10, 20, ..., 100). Applying inclusion-exclusion: 50 + 20 − 10 = 60. Option A overcounts by counting the 10 multiples of 10 twice — the exact error inclusion-exclusion is designed to correct. Option C counts only exclusive membership, which is not what 'or' means in set counting.
Question 2 Multiple Choice
You have sets A, B, and C with |A| = 30, |B| = 25, |C| = 20, |A∩B| = 8, |A∩C| = 6, |B∩C| = 7, and |A∩B∩C| = 3. What is |A ∪ B ∪ C|?
A75 — add all three sets: 30 + 25 + 20
B57 — apply the full formula: 30 + 25 + 20 − 8 − 6 − 7 + 3
C54 — subtract pairwise intersections but forget to add back the triple: 30 + 25 + 20 − 8 − 6 − 7
D60 — add the triple intersection twice to correct for over-subtraction
|A ∪ B ∪ C| = 30 + 25 + 20 − 8 − 6 − 7 + 3 = 57. The most common error is option C — forgetting to add back the triple intersection. The triple intersection |A∩B∩C| was included once in each of the three singleton terms (counted 3 times total), then subtracted once in each of the three pairwise terms (subtracted 3 times). Net count so far: 0. It must be added back once to achieve the correct count of 1.
Question 3 True / False
An element that belongs to all three sets A, B, and C is counted three times by the individual set terms (|A| + |B| + |C|), subtracted three times by the pairwise intersection terms, and must therefore be added back once by the triple intersection term — giving a net count of exactly 1.
TTrue
FFalse
Answer: True
This is the internal accounting of inclusion-exclusion: an element in all three sets contributes +3 from singletons, −3 from pairs (once in each of |A∩B|, |A∩C|, |B∩C|), and +1 from the triple = net 1. A binomial identity guarantees this works for any number of sets: an element in exactly m sets is counted C(m,1) − C(m,2) + C(m,3) − ⋯ = 1.
Question 4 True / False
Inclusion-exclusion mainly applies when counting elements across disjoint sets — if the sets overlap, a different counting method is needed.
TTrue
FFalse
Answer: False
This is exactly backwards. Inclusion-exclusion is specifically designed for overlapping sets. If sets were disjoint, simple addition would suffice — no correction needed. The entire purpose of the principle is to fix the over-counting that occurs when elements belong to multiple sets simultaneously. It is a method for non-disjoint sets, not a restriction to disjoint ones.
Question 5 Short Answer
Why does the inclusion-exclusion formula alternate between adding and subtracting intersection terms, rather than simply subtracting all pairwise overlaps once?
Think about your answer, then reveal below.
Model answer: Simple subtraction of pairwise overlaps over-corrects for elements in three or more sets. An element in three sets gets subtracted three times by pairwise terms, but the correct net removal is only twice (to bring the count from 3 down to 1). So the triple intersection must be added back. For elements in four sets, the pattern continues with further alternations. The alternating sign ensures every element — regardless of how many sets it belongs to — is counted exactly once in the final sum.
This is the core mechanism of the principle. The binomial identity C(m,1) − C(m,2) + C(m,3) − ⋯ ± C(m,m) = 1 for any m ≥ 1 guarantees that any element in exactly m sets gets a net count of 1 after all terms are summed. The alternating signs are not arbitrary — they are the unique correction pattern that achieves this.