Complementary Counting Quiz: Complementary Counting Via Inclusion Exclusion

1) Out of N=200 students, |A|=120 (Art), |B|=90 (Band), and |A ∩ B|=50. How many are in none of A or B?

35

40

45

50

Compute the union with inclusion–exclusion: |A ∪ B| = 120 + 90 − 50 = 160. Complement counts none: 200 − 160 = 40.

Explanation

Compute the union with inclusion–exclusion: |A ∪ B| = 120 + 90 − 50 = 160. Complement counts none: 200 − 160 = 40.

2) For any two sets A and B in a universe of size N, the count in none is N − |A ∪ B|.

True

False

By definition, elements are either in A ∪ B or in its complement; these are disjoint and cover the universe, so |none| = N − |A ∪ B|.

Explanation

By definition, elements are either in A ∪ B or in its complement; these are disjoint and cover the universe, so |none| = N − |A ∪ B|.

3) Out of 100 students, 64 take Math (A), 58 take Science (B), and 30 take both. The number in none is ____.

Use inclusion–exclusion for the union: 64 + 58 − 30 = 92. Then none = 100 − 92 = 8.

Explanation

Use inclusion–exclusion for the union: 64 + 58 − 30 = 92. Then none = 100 − 92 = 8.

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4) In a school of N=150, |A|=70, |B|=65, |C|=60 with |A ∩ B|=30, |A ∩ C|=25, |B ∩ C|=20, and |A ∩ B ∩ C|=12. How many are in none of A,B,C?

12

15

18

22

Three‑set union: 70+65+60 −30−25−20 +12 = 132. Complement gives none = 150 − 132 = 18.

Explanation

Three‑set union: 70+65+60 −30−25−20 +12 = 132. Complement gives none = 150 − 132 = 18.

5) Select all correct statements about complements and inclusion–exclusion.

|none of A or B| = N − (|A| + |B| − |A ∩ B|)

|at least one of A,B| = |A ∪ B|

Exactly one equals |A| + |B| − 2|A ∩ B| (two sets)

If A,B,C are given, then |A ∪ B ∪ C| = |A|+|B|+|C| − |A∩B| − |A∩C| − |B∩C| + |A∩B∩C|

|none| + |A ∪ B| = N

A and E are the complement identity with two sets. B is the definition of “at least one.” D is the three‑set inclusion–exclusion formula. C is false for “exactly one” in two sets; the correct formula is |A\B| + |B\A| = |A| + |B| − 2|A ∩ B|.

Explanation

A and E are the complement identity with two sets. B is the definition of “at least one.” D is the three‑set inclusion–exclusion formula. C is false for “exactly one” in two sets; the correct formula is |A\B| + |B\A| = |A| + |B| − 2|A ∩ B|.

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6) In a class of 80, 50 are in Chess (A), 40 in Drama (B), and 15 in both. How many are in neither?

3

4

5

10

Union = 50 + 40 − 15 = 75. None = 80 − 75 = 5.

Explanation

Union = 50 + 40 − 15 = 75. None = 80 − 75 = 5.

7) If |A| + |B| > N, then at least one person must be in both A and B.

True

False

If |A| + |B| > N, then |A ∩ B| = |A| + |B| − |A ∪ B| ≥ |A| + |B| − N > 0, so the intersection is nonempty.

Explanation

If |A| + |B| > N, then |A ∩ B| = |A| + |B| − |A ∪ B| ≥ |A| + |B| − N > 0, so the intersection is nonempty.

8) In a group of N=60, |A|=22 and |B|=19 with |A ∩ B|=7. The number in neither A nor B is ____.

Union = 22 + 19 − 7 = 34. None = 60 − 34 = 26.

Explanation

Union = 22 + 19 − 7 = 34. None = 60 − 34 = 26.

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9) Given |A|=40, |B|=35, |C|=25, |A∩B|=12, |A∩C|=10, |B∩C|=9, |A∩B∩C|=5. How many are in at least one of A,B,C?

71

72

74

76

Apply three‑set inclusion–exclusion: 40+35+25 −12−10−9 +5 = 74.

Explanation

Apply three‑set inclusion–exclusion: 40+35+25 −12−10−9 +5 = 74.

10) Out of N=120, suppose |A|=70, |B|=55, and |A ∩ B|=25. Select all correct quantities.

Only‑A = 45

Only‑B = 30

None = 20

At least one = 120

|A ∩ B| = 35

Union = 70 + 55 − 25 = 100 ⇒ none = 20. Only‑A = 70 − 25 = 45; Only‑B = 55 − 25 = 30. The union is 100 (not 120). Intersection is given as 25 (not 35).

Explanation

Union = 70 + 55 − 25 = 100 ⇒ none = 20. Only‑A = 70 − 25 = 45; Only‑B = 55 − 25 = 30. The union is 100 (not 120). Intersection is given as 25 (not 35).

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11) If 30 students are in none and |A ∪ B| = 170, what is the total N?

180

190

200

210

Complement partitions the universe: N = |A ∪ B| + |none| = 170 + 30 = 200.

Explanation

Complement partitions the universe: N = |A ∪ B| + |none| = 170 + 30 = 200.

12) For two sets, |none| = N − |A| − |B| + |A ∩ B|.

True

False

This is N minus the two‑set union |A ∪ B| = |A| + |B| − |A ∩ B|.

Explanation

This is N minus the two‑set union |A ∪ B| = |A| + |B| − |A ∩ B|.

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14) In a school of N=75, 12 students are in none of A or B. How many are in at least one of A or B?

57

60

63

68

At least one = N − none = 75 − 12 = 63.

Explanation

At least one = N − none = 75 − 12 = 63.

15) For three sets A,B,C in a universe of size N, which formulas correctly compute the number in none of A,B,C?

N − (|A| + |B| + |C| − |A∩B| − |A∩C| − |B∩C| + |A∩B∩C|)

N − |A ∪ B ∪ C|

N − (|A| + |B| − |A∩B|)

N − (|A| + |B| + |C| − |A∩B| − |A∩C| − |B∩C|)

N − (|A ∪ B| + |C|)

A and B are equivalent ways to express the complement of the union of three sets. C omits C entirely; D drops the triple‑overlap correction; E double‑counts overlaps.

Explanation

A and B are equivalent ways to express the complement of the union of three sets. C omits C entirely; D drops the triple‑overlap correction; E double‑counts overlaps.

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16) Out of N=200 students, 165 are in at least one of A or B. How many are in none?

30

35

40

45

None = N − (at least one) = 200 − 165 = 35.

Explanation

None = N − (at least one) = 200 − 165 = 35.

17) If no one is in none (i.e., |none|=0), then every element lies in A ∪ B ∪ C.

True

False

|none|=0 means the complement of the union is empty, so the union equals the entire universe.

Explanation

|none|=0 means the complement of the union is empty, so the union equals the entire universe.

18) In a group of N=300, |A|=140, |B|=120, |C|=110, |A∩B|=60, |A∩C|=50, |B∩C|=40, |A∩B∩C|=30. The number in none is ____.

Union = 140+120+110 −60−50−40 +30 = 250. None = 300 − 250 = 50.

Explanation

Union = 140+120+110 −60−50−40 +30 = 250. None = 300 − 250 = 50.

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19) If N=220 and 180 students are in at least one of A,B,C, how many are in none?

30

35

40

45

None = N − (at least one) = 220 − 180 = 40.

Explanation

None = N − (at least one) = 220 − 180 = 40.

20) Select all correct general statements about complementary counting with inclusion–exclusion.

Counting the complement is often simpler than counting the union directly

When using complements you can safely ignore overlaps

For two sets, |none| = N − (|A| + |B| − |A∩B|)

If A and B are disjoint, then |none| = N − |A| − |B|

Knowing |none| and N determines |A ∪ B|

A is a common strategy. C and D are the two‑set complement formulas. E follows from |A ∪ B| = N − |none|. B is false because overlaps are the key reason inclusion–exclusion is needed.

Explanation

A is a common strategy. C and D are the two‑set complement formulas. E follows from |A ∪ B| = N − |none|. B is false because overlaps are the key reason inclusion–exclusion is needed.

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Complementary Counting Quiz: Complementary Counting via Inclusion Exclusion