Adding Two Overlapping Groups Counts the Middle Part Twice

Because anyone in both groups gets counted twice.

In a class of $40$, suppose $24$ play cricket and $18$ play football, with $6$ playing both. Adding gives $24+18=42$ — which is more than the class contains.

The error is exactly the $6$ who play both. They were counted once among the cricketers and again among the footballers, so the true number playing at least one game is

$$24+18-6=36$$

Subtracting the overlap once fixes it. That single correction is the whole of the union formula, and this page covers the second part of the ICSE Class 8 Mathematics chapter on sets: the operations, Venn diagrams, the laws they obey, and counting problems.

Three operations, each answering a different question.

Union $A\cup B$ — the set of elements in $A$ or $B$ or both. Read as A union B.

Intersection $A\cap B$ — the set of elements in both $A$ and $B$. Read as A intersection B.

Difference $A-B$ — the set of elements in $A$ but not in $B$.

Worked example. Let $A=\{1,2,3,4,5\}$ and $B=\{4,5,6,7\}$.

$$A\cup B=\{1,2,3,4,5,6,7\},n(A\cup B)=7$$

$$A\cap B=\{4,5\},n(A\cap B)=2$$

$$A-B=\{1,2,3\}$$

$$B-A=\{6,7\}$$

Notice that the **common elements $4$ and $5$ appear only once in the union, because a set never repeats an element.

Difference is not commutative.** $A-B=\{1,2,3\}$ while $B-A=\{6,7\}$, and these are quite different sets. Union and intersection are commutative, so $A\cup B=B\cup A$ and $A\cap B=B\cap A$ — but the order matters for difference, exactly as it does for subtraction of numbers.

Disjoint and overlapping sets:

- Disjoint sets have nothing in common, so $A\cap B=\varnothing$. For example $\{1,2,3\}$ and $\{7,8\}$.
- Overlapping sets have at least one common element, so $A\cap B\ne \varnothing$. The $A$ and $B$ above are overlapping.

Why disjointness simplifies counting. For disjoint sets the overlap is zero, so $n(A\cup B)=n(A)+n(B)$ with nothing to subtract. Adding group sizes is safe only when the groups are disjoint — which is precisely the condition the opening example failed.

Two results worth knowing. $A\cup A=A$ and $A\cap A=A$; and if $A\subseteq B$ then $A\cup B=B$ and $A\cap B=A$. Both follow directly from the definitions and are quick one-mark answers.

Draw the universal set as a rectangle, and each set inside it as a circle. The way the circles lie shows the relationship.

For two sets:

- Overlapping circles show sets with common elements. The diagram then has four regions — only $A$, only $B$, both, and neither.
- Separate circles show disjoint sets, with no both region at all.
- One circle inside another shows a subset, so $A\subseteq B$.

Reading the four regions of an overlapping pair. Using $A=\{1,2,3,4,5\}$ and $B=\{4,5,6,7\}$ inside $\xi =\{1,\dots ,10\}$:

- **Only $A$**, the region $A-B$: $\{1,2,3\}$
- Both, the region $A\cap B$: $\{4,5\}$
- **Only $B$**, the region $B-A$: $\{6,7\}$
- Neither, outside both circles: $\{8,9,10\}$, which is $(A\cup B{)}^{\prime }$

Every element of $\xi$ sits in exactly one of those four regions, so the four counts must add to $n(\xi )$:

$$3+2+2+3=10$$

That check is the single most useful thing a Venn diagram gives you.

Shading the operations. To show an operation, shade the regions it contains:

- $A\cup B$ — shade both circles entirely
- $A\cap B$ — shade only the overlap
- $A-B$ — shade the part of $A$ outside $B$
- $A^{\prime }$ — shade everything outside $A$, including the part of $B$ outside $A$
- $(A\cup B{)}^{\prime }$ — shade only the region outside both circles

For three sets, three mutually overlapping circles give eight regions — one for each combination of in-or-out across the three sets, which is $2^3=8$, the same count as the subsets of the previous part.

How to fill a three-set diagram. Always start with the centre, the region belonging to all three, and work outwards. If you start at the edges you will double-count the middle, because each outer figure given in a question usually includes the centre.

What a Venn diagram is really for. It converts a counting problem into a picture in which every person or object is in exactly one place. Once the regions are filled, every question — how many play only cricket, how many play neither — is read straight off, with no formula needed.

Work out each side separately and completely, then state that they are equal. Verifying means doing both halves, not arguing from the shape of the formula.

De Morgan's laws:

$$(A\cup B{)}^{\prime }=A^{\prime }\cap B^{\prime }$$

$$(A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }$$

In words: the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements. Each law swaps union and intersection as the complement moves inside the bracket.

Verifying the first law. Let $\xi =\{1,2,\dots ,10\}$, $A=\{1,2,3,4,5\}$ and $B=\{4,5,6,7\}$.

Left-hand side:

$$A\cup B=\{1,2,3,4,5,6,7\}\Rightarrow (A\cup B{)}^{\prime }=\{8,9,10\}$$

Right-hand side:

$$A^{\prime }=\{6,7,8,9,10\},B^{\prime }=\{1,2,3,8,9,10\}$$

$$A^{\prime }\cap B^{\prime }=\{8,9,10\}$$

Both sides give $\{8,9,10\}$, so the law is verified.

Verifying the second law, with the same sets:

$$A\cap B=\{4,5\}\Rightarrow (A\cap B{)}^{\prime }=\{1,2,3,6,7,8,9,10\}$$

$$A^{\prime }\cup B^{\prime }=\{6,7,8,9,10\}\cup \{1,2,3,8,9,10\}=\{1,2,3,6,7,8,9,10\}$$

Equal again, so the second law is verified.

Why the operation flips. *Not in ($A$ or $B$) means not in $A$ and not in $B$ — you have to be outside both. And not in ($A$ and $B$) means not in $A$ or not in $B$* — missing from either one is enough. The everyday logic and the set algebra say the same thing, which is why the laws are worth understanding rather than memorising.

The distributive property of union over intersection:

$$A\cup (B\cap C)=(A\cup B)\cap (A\cup C)$$

Verifying it. Let $A=\{1,2,3\}$, $B=\{2,3,4\}$ and $C=\{3,4,5\}$.

Left-hand side — bracket first:

$$B\cap C=\{3,4\}\Rightarrow A\cup (B\cap C)=\{1,2,3,4\}$$

Right-hand side — each bracket, then intersect:

$$A\cup B=\{1,2,3,4\},A\cup C=\{1,2,3,4,5\}$$

$$(A\cup B)\cap (A\cup C)=\{1,2,3,4\}$$

Both sides give $\{1,2,3,4\}$. Verified.

The companion law, which works the same way with the operations exchanged:

$$A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$$

What makes set algebra unlike ordinary algebra. With numbers, multiplication distributes over addition but addition does not distribute over multiplication — $a+(b\times c)$ is not $(a+b)\times (a+c)$. With sets, both distributive laws hold, so union and intersection are on an equal footing in a way that $+$ and $\times$ never are.