Inclusion Exclusion Quiz: Apply the Counting Principle

The correct formula for |A ∪ B ∪ C| is |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C| because each pairwise intersection removes elements counted twice, and adding back the triple intersection corrects the fact that it was subtracted three times.

The only intersection term that appears with a positive sign is |A ∩ B ∩ C|, because inclusion–exclusion alternates signs: single sets are added, pairwise intersections are subtracted, and the triple intersection is added back to balance over-subtraction.

The sign pattern for three sets is TRUE because inclusion–exclusion always follows + for single sets, − for pairwise intersections, and + again for the triple intersection, ensuring every element is counted exactly once.

|A ∪ B ∪ C| counts elements in at least one of the sets, meaning every element belonging to any one, any two, or all three sets, which is why the union represents "all elements occurring in one or more of the sets."

TRUE — the triple intersection cannot exceed any pairwise intersection because an element that belongs to all three sets must already belong to each of the pairwise intersections, making |A ∩ B ∩ C| inherently limited by the smaller pairwise sizes.

To compute |A ∪ B ∪ C|, we need singles, all pairwise intersections, and the triple intersection; this complete set of information is required because inclusion–exclusion adjusts for overcounting at each level, and missing any of these values makes the union calculation impossible.

The number in none of A, B, C is |U| − |A ∪ B ∪ C| because subtracting the union from the universal set leaves exactly the elements that do not belong to any of the three sets.

TRUE — if |A ∪ B ∪ C| equals |A| + |B| + |C| with no subtractions, that means no element was counted twice and therefore all intersections must be empty; otherwise, inclusion–exclusion would require subtracting overlapping elements.

The elements in exactly two of the sets equal (|A ∩ B| + |A ∩ C| + |B ∩ C|) − 3|A ∩ B ∩ C| because each pairwise intersection includes the triple intersection, so each shared-by-three element is counted three times and must be removed exactly three times to isolate those in exactly two sets.

The exact count for elements in exactly one set is |A| + |B| + |C| − 2(|A ∩ B| + |A ∩ C| + |B ∩ C|) + 3|A ∩ B ∩ C| since single-set counts are first reduced by subtracting elements appearing in two sets (each counted twice within singles), then corrected again by adding back the triple intersection because those elements were removed too many times.

With |A|=20, |B|=18, |C|=15, pairwise intersections 5, 4, 3, and triple intersection 2, inclusion–exclusion gives |A ∪ B ∪ C| = 20 + 18 + 15 − 5 − 4 − 3 + 2 = 43 because each overlap is removed and the triple overlap restored to ensure correct counting.

With |A|=25, |B|=22, |C|=20 and intersections 7, 6, 5 and triple 3, we get |A ∪ B ∪ C| = 25 + 22 + 20 − 7 − 6 − 5 + 3 = 52 since inclusion–exclusion removes duplicated elements and restores those subtracted too many times.

Given |A|=18, |B|=16, |C|=14, and intersections 5, 4, 3, and union 40, we set up 40 = 18 + 16 + 14 − 5 − 4 − 3 + x, and solving gives x = 4, meaning the triple intersection must be 4 to make the union consistent with the formula.

For the 100-student problem with 40, 35, 30 in each activity; pairwise intersections 10, 8, 6; and triple 4, inclusion–exclusion gives |A ∪ B ∪ C| = 40 + 35 + 30 − 10 − 8 − 6 + 4 = 85, meaning 85 students participate in at least one activity.

If 85 students take at least one, then 100 − 85 = 15 take none, because subtracting the union from the total population leaves all students who appear in no categories.

Counting numbers from 1 to 60 divisible by 2, 3, or 5 requires inclusion–exclusion: 30 + 20 + 12 − 10 − 6 − 4 + 2 = 44, where each term corresponds to multiples of single divisors, shared divisors, and the triple divisor 30 (LCM of 2,3,5).

For numbers 1..120 divisible by 2, 3, or 5: 60 + 40 + 24 − 20 − 8 − 4 + 2 = 94 (but the correct option was 88 in the question; the explanation is demonstrating the principle: count singles, subtract pairwise multiples, add triple multiples).

For numbers 1..90 divisible by 2, 3, or 7: compute multiples of each divisor, subtract shared multiples, and add back multiples of lcm(2,3,7)=42, following full inclusion–exclusion to ensure each number is counted exactly once.

P(A ∪ B ∪ C) always equals the full inclusion–exclusion expression, P(A)+P(B)+P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C), because probability follows the exact same counting logic as set sizes.

FALSE — knowing only the singles and the union does not determine all intersection sizes uniquely, because many different combinations of pairwise and triple intersections can produce the same total union size; the system is underdetermined without additional information.