Let $U$ be the universal set for sets $A$ and $B$ such that $n(A) = 200$, $n(B) = 300$ and $n(A \cap B) = 100$. Then, $n(A' \cap B') = 300$, provided $n(U)$ is equal to: (i) 600 (ii) 700 (iii) 800 (iv) 900

> [!TLDR] > Use $n(A'\cap B') = n(U) - n(A\cup B)$ together with the union formula to solve for $n(U)$. > [!FORMULA] > $n(A\cup B) = n(A) + n(B) - n(A\cap B)$; and by De Morgan's law, $A'\cap B' = (A\cup B)'$, so $n(A'\cap B') = n(U) - n(A\cup B)$. 1. Given: $n(A)=200$, $n(B)=300$, $n(A\cap B)=100$, and $n(A'\cap B')=300$. 2. Compute the union: $n(A\cup B)=n(A)+n(B)-n(A\cap B)=200+300-100=400$. 3. Since $A'\cap B'=(A\cup B)'$ (De Morgan's law), $n(A'\cap B')=n(U)-n(A\cup B)$. 4. Substitute: $300=n(U)-400$. …

Let U be the universal set for sets A and B such that n(A) = 200, n(B) = 300 and n(A ∩ B) = 100. Then, n(A' ∩

Worked Examples · Example 19

U

(i) 600

(ii) 700

(iii) 800

(iv) 900

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Concept understanding — Practical Problems on Operations on Sets

Practical Problems on Operations on Sets

You already know what a set is — a collection of distinct objects. And you know the basic operations: union ( $A \cup B$ ), intersection ( $A \cap B$ ), difference ( $A - B$ ), and complement ( $A'$ ). The question is: why do we need to solve problems with them? Because in real life, data rarely comes neatly labelled as "these are the elements of set A". Instead, you get a paragraph about a survey, or a group of people, or a list of items with overlapping properties. Your job is to translate that messy reality into set language, apply the operations, and extract the answer.

The core idea: counting with overlaps

Imagine a class of 30 students. 18 play cricket, 15 play football, and 10 play both. How many play at least one sport? How many play neither?

If you simply add 18 + 15, you get 33 — but there are only 30 students. The mistake is that the 10 students who play both have been counted twice. The union formula fixes this:

n(A \cup B) = n(A) + n(B) - n(A \cap B)

Here, $n(A \cup B) = 18 + 15 - 10 = 23$ . So 23 students play at least one sport. The remaining $30 - 23 = 7$ play neither.

Tip

Whenever you see "both" or "and" in a problem, that's the intersection. Whenever you see "at least one" or "either ... or", that's the union. The word "neither" points to the complement of the union.

The general formula for three sets

When three sets overlap, the same logic extends. For sets $A$ , $B$ , $C$ :

n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)

Why the plus at the end? Because when you subtract all three pairwise intersections, the triple intersection (the region common to all three) gets subtracted three times — but it was originally added three times. So it's been removed entirely. Adding it back once corrects that.

Watch out

A common mistake is to forget the $+ n(A \cap B \cap C)$ term. Without it, the triple-overlap region is missing from the count.

How to approach any problem

Identify the universal set — the whole group under consideration. Its size is $n(U)$ .
Define your sets clearly. Usually the problem gives you categories: "people who like tea", "people who like coffee", etc.
Extract the given numbers — total, individual set sizes, pairwise intersections, triple intersection.
Decide what is being asked — union? complement? exactly one? exactly two?
Apply the appropriate formula or draw a Venn diagram.

Venn diagrams: your visual tool

For problems with two or three sets, a Venn diagram is often faster than memorising formulas. Draw three overlapping circles inside a rectangle (the universal set). Label each region. Fill in known numbers from the innermost region outward — start with the triple intersection, then the pairwise overlaps (subtracting the triple), then the individual sets (subtracting the overlaps). Finally, the region outside all circles is $n(U) - n(A \cup B \cup C)$ . …

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Use $n(A'\cap B') = n(U) - n(A\cup B)$ together with the union formula to solve for $n(U)$ .

n(A\cup B) = n(A) + n(B) - n(A\cap B)

Given: $n(A)=200$ , $n(B)=300$ , $n(A\cap B)=100$ , and $n(A'\cap B')=300$ .
Compute the union: $n(A\cup B)=n(A)+n(B)-n(A\cap B)=200+300-100=400$ .
Since $A'\cap B'=(A\cup B)'$ (De Morgan's law), $n(A'\cap B')=n(U)-n(A\cup B)$ .
Substitute: $300=n(U)-400$ . …

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Q.Let UUU be the universal set for sets AAA and BBB such that n(A)=200n(A) = 200n(A)=200, n(B)=300n(B) = 300n(B)=300 and n(A∩B)=100n(A \cap B) = 100n(A∩B)=100. Then, n(A′∩B′)=300n(A' \cap B') = 300n(A′∩B′)=300, provided n(U)n(U)n(U) is equal to:

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Q.Let $U$ be the universal set for sets $A$ and $B$ such that $n(A) = 200$ , $n(B) = 300$ and $n(A \cap B) = 100$ . Then, $n(A' \cap B') = 300$ , provided $n(U)$ is equal to:

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