A set is a well-defined collection of distinct objects. "Well-defined" means there is no ambiguity — either an element belongs to the set or it does not. A group of tall students is not a set (subjective). A group of students who scored above 60 marks is a set (objective).
Think of a set like a closed boundary drawn on paper. Everything inside the boundary belongs to the set. The moment you draw two such boundaries, you can ask: what is inside both? What is inside either one? That is the entire game of set theory at this exam level.
Here is the everyday analogy that makes this stick. Imagine two WhatsApp groups in a colony: Group A has residents who own a car, Group B has residents who own a bike. Some residents appear in both groups — they own both. Now if someone asks "how many unique residents are on these lists?" you cannot just add the two counts, because you would double-count the people in both groups. You subtract the overlap once. That subtraction is the inclusion-exclusion principle, and it is the one formula UP Police Constable sets questions on.
The notation to know:
n(A) — number of elements in set An(B) — number of elements in set Bn(A ∩ B) — elements common to both A and B (intersection — the overlap)n(A ∪ B) — elements in A or B or both (union — the total covered)n(U) — total elements in the universal set (everyone surveyed)n(U) − n(A ∪ B)
Why do you subtract the intersection? Because every element in A ∩ B gets counted once in n(A) and once again in n(B). So n(A) + n(B) overcounts the overlap by exactly one. You subtract it once to correct that.
This is the only derivation worth memorising at this level. Everything else is a rearrangement of this one identity.
The exam rarely gives you exactly what you need. It gives you three out of four quantities and asks for the fourth. Here are all four forms:
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)n(A ∩ B) = n(A) + n(B) − n(A ∪ B)n(A) − n(A ∩ B)n(U) − n(A ∪ B)Rearrangement 4 is the most frequently tested trap. The question gives you a class of 40 students, tells you how many play cricket and tennis, and asks "how many play neither?" You first find the union (those who play at least one), then subtract from the total.
Draw two overlapping circles. Label three regions:
n(A) − n(A ∩ B)n(A ∩ B)n(B) − n(A ∩ B)n(U) − n(A ∪ B)Always fill Region II first (the intersection, which is usually given). Then fill Region I and III. Sum of all four regions = n(U).
This labelling approach turns every set-theory question into a fill-in-the-blank exercise. You stop thinking in abstract formulas and start reading numbers off a diagram.
For completeness — if a question ever involves three sets A, B, C:
UP Police Constable has not tested three-set problems in recent years based on available PYQs, so do not spend time here if you are short on prep time. Know the formula exists; master the two-set version cold.
A set P is a subset of Q (written P ⊆ Q) if every element of P is also in Q. A set P is a proper subset (P ⊂ Q) if P is a subset of Q but P ≠ Q.
The number of subsets of a set with n elements is 2ⁿ. This fact appears occasionally in definition-type MCQs. For a set with 3 elements: 2³ = 8 subsets.
Every time a set question gives you n(A), n(B), and n(A ∩ B) and asks for n(A ∪ B), your hand should automatically write: Add the two, subtract the overlap. No thinking required.
Pattern: n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Micro-example: n(A) = 28, n(B) = 35, n(A ∩ B) = 12 → 28 + 35 − 12 = 51. Done.
Standard method (re-reading question + setting up from scratch): 40 seconds. This reflex: 8 seconds. The formula fires before you finish reading the last number.
"Neither" questions add a final step that many candidates miss. They find the union and then stop — that gives "at least one", not "neither".
Rule: Neither = n(U) − n(A ∪ B)
Micro-example: Class of 40, cricket = 15, tennis = 15, both = 5. Union = 15 + 15 − 5 = 25. Neither = 40 − 25 = 15.
Without this explicit reflex, candidates write 25 and pick the wrong option. With it: two-step, under 15 seconds.
If you are confused about what is being asked, skip straight to drawing the Venn diagram and labelling the three inner regions with actual numbers.
Region II (intersection) = given. Region I = n(A) − Region II. Region III = n(B) − Region II.
Sum of Region I + II + III = n(A ∪ B).
This converts a formula problem into a counting problem. Candidates who draw the Venn diagram make near-zero errors. Candidates who try to hold everything in their head make frequent sign errors. Drawing takes 10 extra seconds but saves 30 seconds of re-checking.
Sometimes the question gives you n(A ∪ B) and asks for n(A ∩ B). Use the same formula in reverse:
n(A ∩ B) = n(A) + n(B) − n(A ∪ B)
Micro-example: n(A) = 30, n(B) = 25, n(A ∪ B) = 45 → intersection = 30 + 25 − 45 = 10.
Standard approach (re-deriving the formula from scratch): 5 steps. Reverse substitution: 1 step. Most candidates do not recognise when to use the reverse gear and waste 30+ seconds.
Read the question and identify which of the four quantities is missing: union, intersection, "only A", or "neither".
Step 1 — Extract: note down n(A), n(B), n(A ∩ B), n(U) from the question. Circle what is missing.
Step 2 — If asked for union: n(A) + n(B) − n(A ∩ B). Done.
Step 3 — If asked for intersection: n(A) + n(B) − n(A ∪ B). Done.
Step 4 — If asked for "neither": First compute union (Step 2), then subtract from n(U).
Step 5 — If asked for "only A" or "only B": Subtract intersection from the respective set's count.
Decision rule: Any question with the words "at least one" is asking for the union. Any question with "neither" or "none of the above" is asking for n(U) − union. Do not confuse these two.
Cross-check: Region I + Region II + Region III + Neither must equal n(U). If it does not, you made an arithmetic error.
Why this question: This is the cleanest, most direct test of the inclusion-exclusion formula. No tricks, no extra steps — just the base formula.
Solving path: n(A) = 28, n(B) = 35, n(A ∩ B) = 12. Apply: 28 + 35 − 12 = 51. The correct answer is option (c). Do not add 28 + 35 = 63 and stop there — that is the most common wrong answer.
Why this question: The word "at least one" is code for "find the union". This question tests whether you recognise that trigger word.
Solving path: n(cold drinks) = 140, n(milkshakes) = 120, n(both) = 80. "At least one" = union = 140 + 120 − 80 = 180. The total surveyed (200) is irrelevant here because the question asks only for the union, not "neither". Do not subtract 180 from 200 — that would answer a different question. Answer: 180.
Why this question: Another direct union question, but with smaller numbers to confirm the formula works the same way regardless of scale.
Solving path: n(A) = 20, n(B) = 18, n(A ∩ B) = 10. Union = 20 + 18 − 10 = 28. Answer: 28. The trap option here is 38 (forgetting to subtract the intersection) — that is what happens when you just add 20 + 18.
Why this question: This is the "neither" variant. It requires a two-step process and is where the most marks are lost in set theory.
Solving path: Total students = 40. Cricket = 15, Tennis = 15, Both = 5. Step 1 — at least one sport: 15 + 15 − 5 = 25. Step 2 — neither: 40 − 25 = 15. Answer: 15. The wrong answer candidates write is 25 — they solve Step 1 correctly but confuse "at least one" with the question being answered. The question asks for "neither", so you must do Step 2.
Adding without subtracting the intersection. Writing n(A ∪ B) = n(A) + n(B) is the single most common error. The intersection is always subtracted once. If the intersection is zero (disjoint sets), n(A ∩ B) = 0 and then the formula still holds — you just subtract zero.
Confusing "at least one" with "exactly one". "At least one" = union = n(A) + n(B) − n(A ∩ B). "Exactly one" = only A + only B = n(A) + n(B) − 2 × n(A ∩ B). These are different numbers, and the exam sometimes offers both as answer choices to trap you.
Stopping before computing "neither". Candidates correctly compute the union (say, 25) and then select 25 as the answer when the question asked for "neither out of 40". Always re-read the question to see if a final subtraction from the total is needed.
Using n(U) when the question does not give a universal set. Not every set theory question involves a total population. If only two sets and their intersection are given, the union is the only answer possible — do not invent a total.
Double-checking with the wrong total. After computing the Venn diagram regions, students sometimes check Region I + Region II + Region III against n(U) directly, forgetting to add the "neither" region. The correct check is: Region I + Region II + Region III + Neither = n(U).
Misreading n(A ∩ B) as n(A ∪ B). In Hindi-medium backgrounds, the symbols ∩ (intersection, cut) and ∪ (union, cup) look similar under time pressure. Remember: ∩ looks like an arch/bridge (connect = both), ∪ looks like a cup that collects everything (union = either). Thirty seconds spent confirming which symbol is which saves 4 marks.