Inclusion–Exclusion For Two Sets

1) Which formulas correctly represent the size of the union of two finite sets A and B?

|A ∪ B| = |A| + |B|

|A ∪ B| = |A| + |B| − |A ∩ B|

|A ∪ B| = |A| − |B| + |A ∩ B|

|A ∪ B| = |A ∩ B|

|A ∪ B| = |A| + |B| − |A ∩ B| is the correct formula because adding |A| and |B| counts every element of A and every element of B but counts shared elements twice, so subtracting the size of the intersection once removes that double count and produces the true size of the union.

Explanation

|A ∪ B| = |A| + |B| − |A ∩ B| is the correct formula because adding |A| and |B| counts every element of A and every element of B but counts shared elements twice, so subtracting the size of the intersection once removes that double count and produces the true size of the union.

2) In which situations would double-counting occur if we added |A| + |B| directly?

Counting students who like both music and drama

Counting only those who like exactly one activity

Adding disjoint categories

Including overlapping members twice

Double-counting occurs whenever an element belongs to both sets, such as students who like both music and drama being counted once in the music group and once in the drama group, so adding |A| + |B| counts those shared students twice unless the intersection is subtracted.

Explanation

Double-counting occurs whenever an element belongs to both sets, such as students who like both music and drama being counted once in the music group and once in the drama group, so adding |A| + |B| counts those shared students twice unless the intersection is subtracted.

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3) Which pair of sets is disjoint?

Even numbers and Odd numbers

Multiples of 2 and Multiples of 4

Prime numbers and Composite numbers (≥2)

Even numbers and Multiples of 3

Two sets are disjoint when they share no common elements—their intersection is the empty set (∅).
A) Even ∩ Odd = ∅, since no number can be both even and odd.
C) Prime ∩ Composite = ∅ for integers ≥ 2, because a number cannot be both prime and composite.
B) Multiples of 2 and 4 overlap (e.g., 4, 8, 12 …), so not disjoint.
D) Even and Multiples of 3 overlap at numbers like 6, 12 …, so not disjoint.

Explanation

Two sets are disjoint when they share no common elements—their intersection is the empty set (∅).
A) Even ∩ Odd = ∅, since no number can be both even and odd.
C) Prime ∩ Composite = ∅ for integers ≥ 2, because a number cannot be both prime and composite.
B) Multiples of 2 and 4 overlap (e.g., 4, 8, 12 …), so not disjoint.
D) Even and Multiples of 3 overlap at numbers like 6, 12 …, so not disjoint.

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4) If A ⊆ B, then which are true?

|A ∪ B| = |B|

|A ∩ B| = |A|

|A ∪ B| = |A|

|A ∩ B| = |B|

If A ⊆ B, then everything in A is already inside B, so their union simply contains all elements of B giving |A ∪ B| = |B|, while their intersection contains exactly the elements of A giving |A ∩ B| = |A|, and the statements involving |A| for the union or |B| for the intersection would only hold if the two sets were actually equal.

Explanation

If A ⊆ B, then everything in A is already inside B, so their union simply contains all elements of B giving |A ∪ B| = |B|, while their intersection contains exactly the elements of A giving |A ∩ B| = |A|, and the statements involving |A| for the union or |B| for the intersection would only hold if the two sets were actually equal.

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5) Given |A| = 8, |B| = 6, |A ∩ B| = 2, find |A ∪ B|.

10

12

14

8

Using the Inclusion–Exclusion formula |A ∪ B| = |A| + |B| − |A ∩ B| gives 8 + 6 − 2 = 12, which accounts for subtracting the 2 elements shared by both sets that were counted twice when adding 8 and 6.

Explanation

Using the Inclusion–Exclusion formula |A ∪ B| = |A| + |B| − |A ∩ B| gives 8 + 6 − 2 = 12, which accounts for subtracting the 2 elements shared by both sets that were counted twice when adding 8 and 6.

6) When two sets have no common elements, |A ∩ B| = ?

0

1

|A| + |B|

Undefined

If two sets have no elements in common, then their intersection is the empty set ∅ and the size of ∅ is 0, so |A ∩ B| = 0.

Explanation

If two sets have no elements in common, then their intersection is the empty set ∅ and the size of ∅ is 0, so |A ∩ B| = 0.

7) Which are always true about |A ∪ B| = |A| + |B| − |A ∩ B|?

|A ∪ B| ≤ |A| + |B|

|A ∩ B| ≤ |A|

|A ∩ B| ≤ |B|

|A ∩ B| > |A| + |B|

From |A ∪ B| = |A| + |B| − |A ∩ B| we know the union cannot exceed the total |A| + |B| because a non-negative intersection is subtracted, and |A ∩ B| must be less than or equal to both |A| and |B| since the intersection is a subset of each, and therefore cannot exceed either set in size.

Explanation

From |A ∪ B| = |A| + |B| − |A ∩ B| we know the union cannot exceed the total |A| + |B| because a non-negative intersection is subtracted, and |A ∩ B| must be less than or equal to both |A| and |B| since the intersection is a subset of each, and therefore cannot exceed either set in size.

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8) If |A| = 15, |B| = 12, and |A ∪ B| = 20, then |A ∩ B| = ?

7

8

10

15

Using |A ∩ B| = |A| + |B| − |A ∪ B|, we compute 15 + 12 − 20 = 7, meaning 7 elements are counted in both sets because subtracting the union from the total of both individual counts isolates the overlap.

Explanation

Using |A ∩ B| = |A| + |B| − |A ∪ B|, we compute 15 + 12 − 20 = 7, meaning 7 elements are counted in both sets because subtracting the union from the total of both individual counts isolates the overlap.

9) The Inclusion–Exclusion Principle applies only when sets are disjoint.

True

False

This statement is false because Inclusion–Exclusion is specifically required when sets overlap to correct double-counting, but when sets are disjoint the intersection is zero and the formula simplifies to simple addition without needing any correction.

Explanation

This statement is false because Inclusion–Exclusion is specifically required when sets overlap to correct double-counting, but when sets are disjoint the intersection is zero and the formula simplifies to simple addition without needing any correction.

10) For disjoint sets, |A ∪ B| = |A| + |B|.

True

False

This statement is true because disjoint sets have |A ∩ B| = 0, so substituting into the formula gives |A ∪ B| = |A| + |B| − 0, meaning the union is exactly the sum of the two separate sizes.

Explanation

This statement is true because disjoint sets have |A ∩ B| = 0, so substituting into the formula gives |A ∪ B| = |A| + |B| − 0, meaning the union is exactly the sum of the two separate sizes.

11) If |A| = 10, |B| = 12, |A ∩ B| = 4, then |A ∪ B| = 22.

True

False

The statement is false because the correct computation is |A ∪ B| = 10 + 12 − 4 = 18, and saying 22 would mean ignoring the overlap of 4 elements, leading to double-counting them.

Explanation

The statement is false because the correct computation is |A ∪ B| = 10 + 12 − 4 = 18, and saying 22 would mean ignoring the overlap of 4 elements, leading to double-counting them.

12) The same formula works for probabilities: P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

True

False

This statement is true because probability behaves like a normalized version of set size, so the probability of A or B occurring is P(A) + P(B) minus P(A ∩ B) to remove the probability of outcomes counted in both events.

Explanation

This statement is true because probability behaves like a normalized version of set size, so the probability of A or B occurring is P(A) + P(B) minus P(A ∩ B) to remove the probability of outcomes counted in both events.

13) |A ∩ B| can never be larger than |A| or |B|.

True

False

This statement is true because A ∩ B is a subset of A and also a subset of B, and a subset can never contain more elements than the set it belongs to, so |A ∩ B| ≤ |A| and ≤ |B| always holds.

Explanation

This statement is true because A ∩ B is a subset of A and also a subset of B, and a subset can never contain more elements than the set it belongs to, so |A ∩ B| ≤ |A| and ≤ |B| always holds.

14) If A = {1,2,3} and B = {3,4,5}, then |A ∩ B| = 1.

True

False

The statement is true because A = {1, 2, 3} and B = {3, 4, 5} share exactly the element 3, so their intersection is {3} and therefore |A ∩ B| = 1.

Explanation

The statement is true because A = {1, 2, 3} and B = {3, 4, 5} share exactly the element 3, so their intersection is {3} and therefore |A ∩ B| = 1.

15) The principle of ____ and ____ prevents double-counting when combining sets.

The missing words are Inclusion and Exclusion because we first include all elements of both sets by adding their sizes and then exclude the elements counted twice by subtracting the size of the intersection, which prevents double-counting.

Explanation

The missing words are Inclusion and Exclusion because we first include all elements of both sets by adding their sizes and then exclude the elements counted twice by subtracting the size of the intersection, which prevents double-counting.

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16) If |A ∪ B| = 30, |A| = 18, |B| = 20, then |A ∩ B| = ____.

Using |A ∩ B| = |A| + |B| − |A ∪ B|, we compute 18 + 20 − 30 = 8, meaning that 8 elements must be common to both sets because that is the amount necessary to reduce the union count down to 30 after adding the individual set sizes.

Explanation

Using |A ∩ B| = |A| + |B| − |A ∪ B|, we compute 18 + 20 − 30 = 8, meaning that 8 elements must be common to both sets because that is the amount necessary to reduce the union count down to 30 after adding the individual set sizes.

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17) For disjoint sets, |A ∩ B| = ____.

For disjoint sets the intersection contains no elements at all, so |A ∩ B| = 0 because no element belongs to both sets simultaneously.

Explanation

For disjoint sets the intersection contains no elements at all, so |A ∩ B| = 0 because no element belongs to both sets simultaneously.

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18) The region common to both sets is called the ____.

The region common to both sets is called the intersection because it contains exactly the elements that satisfy the condition of belonging to A and B at the same time.

Explanation

The region common to both sets is called the intersection because it contains exactly the elements that satisfy the condition of belonging to A and B at the same time.

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19) If |A| = 9, |B| = 5, and |A ∪ B| = 11, then |A ∩ B| = ____.

Using the formula |A ∩ B| = |A| + |B| − |A ∪ B| gives 9 + 5 − 11 = 3, meaning there are 3 elements counted in both sets and subtracting the union isolates that overlap.

Explanation

Using the formula |A ∩ B| = |A| + |B| − |A ∪ B| gives 9 + 5 − 11 = 3, meaning there are 3 elements counted in both sets and subtracting the union isolates that overlap.

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20) The Inclusion–Exclusion Principle ensures that overlapping members are counted exactly ____.

The Inclusion–Exclusion Principle ensures overlapping members are counted exactly once because adding |A| and |B| counts shared elements twice and subtracting |A ∩ B| removes one of those counts, leaving each element represented exactly once.

Explanation

The Inclusion–Exclusion Principle ensures overlapping members are counted exactly once because adding |A| and |B| counts shared elements twice and subtracting |A ∩ B| removes one of those counts, leaving each element represented exactly once.

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Inclusion–Exclusion for Two Sets