Inclusion–Exclusion for Two Sets

The answer is False. The correct computation is union = 10 + 12 - 4 = 18. The value 22 results from ignoring the 4 shared elements and adding without subtracting the overlap, which double-counts those 4 elements.

Adding the two set sizes counts shared elements twice. Subtracting the intersection once removes one of those counts, leaving each shared element counted exactly once. Option A describes the problem before correction. Option C would mean shared elements are excluded entirely. Option D would require subtracting the intersection twice.

A∩B = A + B - union = 9 + 5 - 11 = 3. Option A gives 1, option B gives 2, option D gives 4, none of which correctly rearrange and apply the formula.

The intersection contains exactly the elements belonging to both sets simultaneously. Option A names the region covering all elements in either set. Option B names the elements outside a set. Option D names the relationship where one set is contained within another, not a region itself.

Disjoint sets share no elements so their intersection is the empty set. The size of the empty set is 0. Option B implies one shared element which would mean the sets are not disjoint. Option C gives the union size not the intersection. Option D is incorrect because the result is well-defined as 0.

A∩B = A + B - union = 18 + 20 - 30 = 8. Option A gives 6, option B gives 7, option D gives 9, none of which correctly rearrange and apply the formula.

The Inclusion-Exclusion Principle first includes all elements of both sets by adding their sizes, then excludes elements counted twice by subtracting the intersection. This two-step process ensures each element is counted exactly once. Option A names arithmetic operations not the principle. Options C and D name set operations or general processes unrelated to the principle's name.

The answer is True. The only element belonging to both A and B is 3. The intersection is {3} which contains exactly one element. Checking: 1 and 2 are only in A, and 4 and 5 are only in B, leaving 3 as the sole shared element.

The answer is True. The intersection is a subset of both A and B. A subset cannot contain more elements than the set it belongs to. Therefore the intersection size is at most the size of A and also at most the size of B, making it bounded by the smaller of the two.

The answer is True. Probability behaves like a normalized version of set size. The probability of A or B occurring is found the same way — add the individual probabilities and subtract the probability of both occurring together, which removes the double-count of outcomes belonging to both events.

Adding A and B counts every element in both sets but counts shared elements twice. Subtracting the intersection once removes that double count, giving the true union size. Option A omits the correction for overlap. Option C incorrectly subtracts B and adds the intersection. Option D equates the union with just the intersection, which is only valid when the sets are equal.

The answer is True. Disjoint sets have an intersection of size 0. Substituting into the formula gives union = A + B - 0 = A + B. With no shared elements every element belongs to exactly one set, so the union is simply the combined count.

The answer is False. Inclusion-Exclusion is specifically designed for overlapping sets to correct double-counting. When sets are disjoint the intersection is zero and the formula simplifies to plain addition. Requiring disjoint sets would eliminate the very situations the principle is meant to handle.

A∩B = A + B - union = 15 + 12 - 20 = 7. Option B gives 8, option C gives 10, option D gives 15, none of which correctly rearrange and apply the formula.

Since the intersection is non-negative, subtracting it from A plus B cannot increase the result, so union is at most A plus B, confirming A. The intersection is a subset of A so it cannot exceed A, confirming B. The intersection is also a subset of B so it cannot exceed B, confirming C. Option D is false — the intersection can never exceed either individual set let alone their sum.

If two sets share no elements their intersection is the empty set, and the size of the empty set is 0. Option B gives 1, implying one shared element. Option C gives the sum of both sizes, which is the union of disjoint sets not the intersection. Option D is incorrect because the intersection is well-defined and equals the empty set.

Union = A + B - A∩B = 8 + 6 - 2 = 12. The 2 shared elements are subtracted once to remove the double count. Option A gives 10, option C gives 14, option D gives 8, none of which correctly apply the formula.

If A is contained within B, the union simply equals B since A adds no new elements, confirming A. The intersection equals exactly A since every element of A is in B, confirming B. Option C would require A and B to be the same size. Option D would require A and B to be equal sets.

No number can be both even and odd simultaneously so Even∩Odd is empty, confirming A. No integer greater than or equal to 2 can be both prime and composite so Prime∩Composite is empty, confirming C. Option B: multiples of 4 such as 4, 8, 12 are also multiples of 2, so the sets overlap. Option D: numbers like 6 and 12 are both even and multiples of 3, so they overlap.

When students like both music and drama they are counted once in the music group and again in the drama group, confirming A. Including overlapping members twice is precisely what direct addition does when sets share elements, confirming D. Option B counts only exclusive members so no overlap exists. Option C involves disjoint sets with no shared elements, so no double-counting occurs.