Venn Diagrams

In the standard representation of a Venn diagram, the outer rectangle denotes the universal set U. It provides the complete context or collection of objects under discussion. Circles or other closed curves drawn inside the rectangle represent particular sets such as A and B. The empty set is represented by no enclosed region, while A ∩ B is only the overlapping part of two circles. Therefore the rectangle represents U, making option A correct.

The common or overlapping region of two circles contains exactly those elements that belong to both A and B. This set is called the intersection and is written as A ∩ B. The union A ∪ B includes every region belonging to A or B, including the overlap. A − B contains elements only in A, not in B, and A′ contains elements outside A relative to U. Therefore the common part represents A ∩ B, option B.

The union A ∪ B consists of every element that belongs to A, to B, or to both sets. On a Venn diagram, this means shading the complete area of both circles, including their overlapping region. “Only A” excludes the overlap, and “only B” also excludes it. The region outside the universal set is not part of the diagram’s considered universe. Therefore A ∪ B represents A or B or both, so option C is correct.

The complement A′ is defined relative to the universal set U. It contains every element of U that does not belong to A. In a Venn diagram, A is shown as a circle inside the rectangle U, so A′ is the portion of the rectangle outside the circle A. It is not the interior of A, the overlap with B, or the entire universal set. Hence the correct answer is the region outside A but inside U, option B.

For two finite sets, the inclusion–exclusion formula is n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The intersection is subtracted because its elements have been counted once in n(A) and once again in n(B). Substituting the given values gives 12 + 9 − 4 = 17. Thus the union contains 17 elements, and option A is correct. Adding 12 and 9 without subtraction would double-count the four common elements.

Two sets are called disjoint when they have no common elements. The intersection \(A\cap B\) represents the elements that belong to both \(A\) and \(B\). Since disjoint sets share no element, their intersection is the empty set: \(A\cap B=\varnothing\). In a Venn diagram, the two circles do not overlap. Hence option C is the only correct answer.

The statement \(A\subseteq B\) means that every element of \(A\) is also an element of \(B\). Therefore, the entire region representing \(A\) must lie within the region representing \(B\). In a Venn diagram, this is shown by drawing the circle for \(A\) completely inside the circle for \(B\). Thus, option B is correct; the other placements contradict the definition of a subset.

The region containing elements only in \(A\) consists of elements that belong to \(A\) but do not belong to \(B\). This is the set difference \(A-B\), also written as \(A\cap B'\). The intersection \(A\cap B\) contains common elements, \(B-A\) represents only \(B\), and \(A\cup B\) includes both sets. Therefore, option B is correct.

The set \(B\) consists of two non-overlapping parts: the elements only in \(B\), represented by \(B-A\), and the elements common to both sets, represented by \(A\cap B\). Thus, \(n(B)=n(B-A)+n(A\cap B)=8+4=12\). Therefore, option C is correct. Neither 8 nor 4 represents the entire set \(B\), and 32 is an unjustified product.

The union \(A\cup B\) contains every element that is in \(A\), in \(B\), or in both. Its Venn diagram has exactly three disjoint regions: only \(A\), the overlap \(A\cap B\), and only \(B\). Therefore, \(n(A\cup B)=6+5+4=15\). Option C is correct. Adding only the exclusive regions would omit the overlap, while counting totals separately could double-count it.

The union A ∪ B contains every element that belongs to A, to B, or to both. Therefore, the elements belonging to neither set are outside the union but still inside the universal set U. Subtract the union count from the universal-set count: n(U) − n(A ∪ B) = 40 − 26 = 14. Hence, option A is correct. In a Venn diagram, this is the region inside the rectangle but outside both circles.

A' represents the complement of A, meaning all elements outside A. Similarly, B' contains all elements outside B. Their intersection A' ∩ B' therefore consists of elements outside both circles. By De Morgan’s law, A' ∩ B' = (A ∪ B)'. Thus, the region is neither A nor B, so option C is correct.

For two finite sets, the inclusion–exclusion principle states n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The five common elements are counted once in each set total, so they must be subtracted once to avoid double counting. Hence n(A ∪ B) = 13 + 17 − 5 = 25. Option A is correct; simply adding gives 30 and ignores the overlap.

Let T be the set of tea lovers and C the set of coffee lovers. By the inclusion–exclusion rule, n(T ∪ C) = n(T) + n(C) − n(T ∩ C) = 22 + 18 − 7 = 33. Therefore, the number who like neither is the total minus the union: 40 − 33 = 7. Hence, option A is correct.

The 13 members who like dance include the 4 members who like both music and dance. Therefore, the members who like only dance are found by removing the intersection from the dance set: 13 − 4 = 9. The club total is consistent but is not required for this direct calculation. Hence, option B is correct.

The notation A ∩ B ∩ C means the intersection of all three sets. Therefore, an element belongs to this region only if it is simultaneously in A, in B, and in C. In a three-set Venn diagram, this is the central region where all three circles overlap. It is not a region belonging to just one set or a region outside the sets.

The notation n(A ∩ B ∩ C) gives the number of elements common to A, B, and C simultaneously. In a three-set Venn diagram, the triple intersection is the central region shared by all three circles. Therefore, the number 4 must be written in that central common region, not in any only-one-set region or outside the universal set. Option C is correct.

The two sets contain a common part of 2 elements. Therefore, the elements only in A are n(A) − n(A ∩ B) = 9 − 2 = 7, and the elements only in B are n(B) − n(A ∩ B) = 7 − 2 = 5. Adding these two separate regions gives 7 + 5 = 12, so option B is correct.

The condition A ∩ B = ∅ means that A and B are disjoint, so they have no common elements. Consequently, no subtraction for overlap is needed: n(A ∪ B) = n(A) + n(B) = 6 + 11 = 17. Therefore option C is correct. The value 11 ignores A, 5 is an irrelevant difference, and 66 incorrectly multiplies the cardinalities.

The difference A − B consists of all elements that belong to A but do not belong to B. The complement B′ represents elements outside B, so requiring membership in both A and B′ gives A − B = A ∩ B′. This is the region inside A but outside B, hence option A.