A ∪ B, A ∩ B, A'. n(A ∪ B) = n(A) + n(B) - n(A ∩ B).

Union, Intersection, Complement of Sets — Venn Diagram

Question

In a class of 40 students, 25 play cricket, 20 play football, and 10 play both. Using set theory, find how many students play at least one sport. Also find how many play only cricket and how many play only football.

Solution — Step by Step

Let $C$ = set of cricket players, $F$ = set of football players.

We have $n(C) = 25$ , $n(F) = 20$ , $n(C \cap F) = 10$ , and total students = 40.

The key formula here is:

n(C \cup F) = n(C) + n(F) - n(C \cap F)

Why do we subtract the intersection? Because students who play both sports get counted once in $n(C)$ and once in $n(F)$ — so we’ve double-counted them. Subtracting $n(C \cap F)$ corrects this.

n(C \cup F) = 25 + 20 - 10 = \mathbf{35}

“Only cricket” means cricket players who do NOT play football — that’s $C$ minus the overlap.

n(\text{only } C) = n(C) - n(C \cap F) = 25 - 10 = \mathbf{15}

Same logic for football:

n(\text{only } F) = n(F) - n(C \cap F) = 20 - 10 = \mathbf{10}

The three regions of our Venn diagram should add up to 35:

15 \text{ (only C)} + 10 \text{ (both)} + 10 \text{ (only F)} = 35 \checkmark

Students who play neither sport: $40 - 35 = \mathbf{5}$ .

Why This Works

The union formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ is essentially a counting correction. When we add $n(A)$ and $n(B)$ directly, elements in $A \cap B$ appear in both counts — so we subtract them once to get the true total.

Visually, draw two overlapping circles. The left crescent is “only A”, the right crescent is “only B”, and the overlapping middle is $A \cap B$ . The union covers all three regions. This picture should be your mental model for any such problem.

The complement $A'$ covers everything outside $A$ in the universal set $U$ . So $n(A') = n(U) - n(A)$ . In our problem, “neither sport” is $(C \cup F)'$ , and $n((C \cup F)') = 40 - 35 = 5$ .

Alternative Method

Instead of the formula, we can fill the Venn diagram region by region — a useful approach when the problem gives you more conditions.

Start from the inside out:

Intersection first: 10 students play both — write 10 in the middle.
Only cricket: $25 - 10 = 15$ — write 15 in the left crescent.
Only football: $20 - 10 = 10$ — write 10 in the right crescent.
Total in at least one sport: $15 + 10 + 10 = 35$ .

This approach works especially well in problems with three sets (three overlapping circles), which appear frequently in JEE Main. Master this region-filling method — it scales better than trying to apply a single formula to complex three-set problems.

For three sets: $n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$ . Notice the pattern — subtract pairwise intersections, add back the triple intersection. This is inclusion-exclusion, and it shows up directly in CBSE Class 11 and JEE Main.

Common Mistake

The most frequent error is using $n(A \cup B) = n(A) + n(B)$ without subtracting the intersection. Students who do this get $25 + 20 = 45$ , which already exceeds the total class size of 40 — a clear signal something is wrong. Whenever your union count exceeds $n(U)$ , it means you forgot to subtract $n(A \cap B)$ . Always check: union cannot be larger than the universal set.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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