Q.Show that the set of letters needed to spell “ CA TARACT ” and the set of letters needed to spell “ TRACT” are equal
Concept understanding — Set Difference
Set Difference
The idea in plain words
Imagine two groups of students: those who play cricket (A) and those who play football (B). The set difference A−B (also written A∖B) answers one specific question: "Who plays cricket but NOT football?" You start with everything in A, then remove whatever also happens to be in B.
Set difference is a one-way street: A−B keeps only what's uniquely in A. It has nothing to do with what's uniquely in B.
The precise definition
For two sets A and B:
A−B={x∣x∈A and x∈/B}
Read as: "the set of all x such that x is in A but x is not in B."
Worked example
Let:
A={1,2,3,4,5},B={3,4,5,6,7}
Step 1: Go through each element of A.
Step 2: Keep it only if it is NOT also in B.
- 1∈A, 1∈/B → keep
- 2∈A, 2∈/B → keep
- 3∈A, 3∈B → remove
- 4∈A, 4∈B → remove
- 5∈A, 5∈B → remove
A−B={1,2}
Now compute the other direction:
B−A={6,7}
Notice A−B=B−A — set difference is not commutative.
Key properties
| Property | Statement |
|---|---|
| Not commutative | A−B=B−A in general |
| Difference with itself | A−A=∅ |
| Difference with empty set | A−∅=A, and ∅−A=∅ |
| Difference with universal set | U−A=Ac (the complement of A) |
| Disjoint sets | If A∩B=∅, then A−B=A |
The last property is worth pausing on: if two sets share nothing in common, subtracting one from the other changes nothing — there was nothing to remove.
Set difference vs. complement — the classic mix-up
Students frequently confuse A−B with Ac (complement of A). The difference is what you're comparing against:
- Complement Ac is always relative to the universal set U: everything outside A.
- Difference A−B is relative to whatever second set you name: everything in A that isn't in B.
In fact, complement is just a special case: Ac=U−A.
Set difference vs. symmetric difference
A−B only keeps the "A-only" region. If you want both one-sided regions together (everything in exactly one of the two sets), that's the symmetric difference A△B=(A−B)∪(B−A) — a different, related, but distinct operation.
Why it matters
Set difference shows up constantly in exam problems: "find the elements in A but not in B," Venn-diagram shading questions, and as a building block for symmetric difference and complement. Getting the direction right (A−B vs. B−A) is the single most common source of lost marks.
Takeaway
A−B keeps only what's uniquely in A, throwing away anything shared with B. Always check which set you're subtracting from — the direction changes the answer.
This topic regularly comes up in searches like "set difference formula A minus B" and "set difference vs complement class 11 maths," both grounded in the Sets chapter of the NCERT/CBSE Class 11 Mathematics syllabus. Getting the direction of subtraction right is a classic source of lost marks in board exams and a frequent JEE Main conceptual question.
Why this formula?
Let's break down the definition of a set — not as a formula to memorise, but as a fundamental idea that underpins all of mathematics.
1. What is a Set? (The Core Idea)
A set is a well-defined collection of distinct objects.
The "why" here is about clarity and precision — we need to know exactly what belongs and what does not.
- Well-defined: For any object, we can say yes or no — no ambiguity.
- Distinct: No duplicates — each object appears only once.
Why? Because if we couldn't decide membership, we couldn't do any logical operations. Sets are the building blocks of all mathematical structures.
2. The Key "Formula": Set-Builder Notation
The most common way to define a set is:
S={x∣P(x)}
This reads: "S is the set of all objects x such that property P(x) is true."
Why does this work?
- x is a placeholder for any object.
- P(x) is a logical condition (a predicate) that is either true or false for each x.
- The vertical bar ∣ means "such that".
Example:
A={n∣n∈N,n is even}
Here, P(n) is "n is a natural number and n is even".
Only those n that satisfy both conditions are included.
Why this form? It avoids listing infinitely many elements. It gives a rule — a decision procedure — for membership.
3. The Two Fundamental Properties (Axioms)
Every set definition relies on two intuitive truths:
(a) Extensionality — Two sets are equal if they have the same elements.
A=B⟺(∀x)(x∈A⟺x∈B)
Why? A set is completely determined by its members. There is no other hidden property.
If you know what's inside, you know the set.
(b) Membership — The only relation is ∈ (belongs to).
x∈Sorx∈/S
Why? Because a set is just a container. The only question we can ask is: "Is this object inside?"
4. Why Can't We Just List Everything?
For small sets, listing works:
{1,2,3}
But for infinite sets (like all natural numbers), listing is impossible.
Set-builder notation solves this by giving a rule instead of a list.
Example:
N={n∣n is a positive integer}
This is not a formula to memorise — it's a definition by property.
5. The "Empty Set" — Why It Exists
The empty set ∅ (or {}) is the set with no elements.
∅={x∣x=x}
Why is this allowed?
Because the condition x=x is always false — no object satisfies it.
This is a logical necessity: if we can define a set by a property, we must allow the possibility that nothing satisfies it.
Key insight: The empty set is not "nothing" — it's a set that contains nothing. It's a mathematical object.
6. Summary: The "Why" Behind the Definition
| Concept | Why it's defined this way |
|---|---|
| Set | To have a precise, unambiguous collection — no guesswork. |
| Set-builder | To define infinite or complex sets without listing. |
| Membership (∈) | The only question that matters — is it inside or not? |
| Empty set | Logical completeness — a property may have no objects. |
Final takeaway: The definition of a set is not a formula to plug numbers into. It's a logical framework for saying: "These objects, and only these, belong here." Every formula you see later (union, intersection, complement) builds on this single idea.
Concept: Set Equality via Set Difference
Two sets are equal if and only if they contain exactly the same elements (ignoring repetition and order).
First, identify the distinct letters in each word:
- "CATARACT" uses the letters {C,A,T,R}
- "TRACT" uses the letters {T,R,A,C}
Both sets contain the four letters C, A, T, and R. Even though "CATARACT" has repeated letters (three A's, two T's, two C's), a set records only distinct elements. Similarly, "TRACT" has repeated letters (T appears twice), but the set is {T,R,A,C}.
Since both sets have identical elements, they are equal: {C,A,T,R}={T,R,A,C}.
The two sets are equal because both contain exactly the letters {C,A,T,R}.
Two sets are equal when they contain exactly the same elements (repetition doesn't matter in sets). Both "CATARACT" and "TRACT" require the letters {C,A,T,R}, so the sets are equal.
The heart of this problem lies in understanding what a set is. A set is a collection of distinct objects — each element either belongs or it doesn't, and listing an element multiple times doesn't change the set. When we ask for "the set of letters needed to spell" a word, we're asking: which distinct letters appear at least once?
Let's identify the letters in each word carefully.
-
Find the distinct letters in "CATARACT".
Write out the word: C-A-T-A-R-A-C-T.
Now list each letter that appears, ignoring repetitions:
- C appears (twice, but we count it once)
- A appears (three times, but we count it once)
- T appears (twice, but we count it once)
- R appears (once)
So the set of letters is {C,A,T,R}.
-
Find the distinct letters in "TRACT".
Write out the word: T-R-A-C-T.
List each distinct letter:
- T appears (twice, but we count it once)
- R appears (once)
- A appears (once)
- C appears (once)
So the set of letters is {T,R,A,C}.
-
Compare the two sets.
We have:
Letters in "CATARACT"={C,A,T,R}
Letters in "TRACT"={T,R,A,C}
Sets have no inherent order, so {C,A,T,R} and {T,R,A,C} are the same set — they contain exactly the same four elements.
A common mistake is to think that because "CATARACT" has more letters (8) than "TRACT" (5), the sets must be different. But sets ignore multiplicity: {A,A,A}={A}.
The set of letters needed to spell "CATARACT" is {C,A,T,R}, which equals the set of letters needed to spell "TRACT", namely {T,R,A,C}={C,A,T,R}.
Method: Set Equality via Element Listing
We will use the Listing Method — write both sets explicitly by listing their distinct elements, then compare.
Step 1: Write the first set (from “CATARACT”)
Take the word CATARACT and list each distinct letter (ignore repetitions):
- C, A, T, A, R, A, C, T
- Distinct letters: C, A, T, R
So,
S1={C,A,T,R}
Step 2: Write the second set (from “TRACT”)
Take the word TRACT and list each distinct letter:
- T, R, A, C, T
- Distinct letters: T, R, A, C
So,
S2={T,R,A,C}
Step 3: Compare the two sets
- S1={C,A,T,R}
- S2={T,R,A,C}
Both contain exactly the same four letters. Order does not matter in a set, and repetitions are ignored.
Step 4: Conclusion
Since every element of S1 is in S2 and every element of S2 is in S1, we have:
S1=S2
Therefore, the set of letters needed to spell “CATARACT” equals the set of letters needed to spell “TRACT”.
Here are the most common mistakes students make on this Set Difference / Set Equality problem, along with how to avoid each.
Mistake 1: Counting repeated letters as separate elements
The error:
Students write the set for “CATARACT” as {C,A,T,A,R,A,C,T} and then think the two sets are different because “CATARACT” has more letters.
Why it’s wrong:
A set does not contain duplicates. Each element appears only once. The letter A appears three times in “CATARACT”, but in the set it is listed just once.
How to avoid:
Always write the roster form of a set by listing each distinct letter only once.
- For “CATARACT”: distinct letters are C,A,T,R → set = {C,A,T,R}
- For “TRACT”: distinct letters are T,R,A,C,T → set = {T,R,A,C}
Both sets are {A,C,R,T} — equal.
Mistake 2: Confusing “set of letters” with “multiset” or “word”
The error:
Students think the order or frequency matters (e.g., “CATARACT” has 3 A’s, “TRACT” has 1 A, so they can’t be equal).
Why it’s wrong:
A set is unordered and has no multiplicity. The question explicitly says “set of letters”, not “list of letters” or “multiset”.
How to avoid:
Remember the definition:
A set is a collection of distinct objects, with no order.
If the problem meant frequency, it would say “multiset” or “bag of letters”.
Mistake 3: Forgetting to check both subset directions
The error:
Students only check that every letter in “TRACT” is in “CATARACT” (true), but don’t check the reverse.
Why it’s wrong:
Set equality requires A⊆B and B⊆A. If you only check one direction, you might miss that “CATARACT” has a letter not in “TRACT” (it doesn’t, but the habit is important).
How to avoid:
Always do the two-way check:
-
Forward: Is every letter of “TRACT” in “CATARACT”?
- T ✓, R ✓, A ✓, C ✓ → Yes.
-
Backward: Is every letter of “CATARACT” in “TRACT”?
- C ✓, A ✓, T ✓, R ✓ → Yes.
Since both hold, the sets are equal.
Mistake 4: Writing the set in the wrong order and thinking order matters
The error:
A student writes {C,A,T,R} for “CATARACT” and {T,R,A,C} for “TRACT”, then says they are different because the order is different.
Why it’s wrong:
Sets are unordered. {C,A,T,R}={T,R,A,C}.
How to avoid:
When comparing two sets, sort the elements alphabetically (or by any fixed rule) before comparing. This makes equality obvious.
- Sorted: both become {A,C,R,T}.
Mistake 5: Including spaces or punctuation as “letters”
The error:
The word “CATARACT” is written with a space in the problem: “CA TARACT”. Some students include the space as an element.
Why it’s wrong:
The problem says “letters needed to spell” — spaces are not letters.
How to avoid:
Ignore spaces, hyphens, apostrophes, etc. Only consider the alphabet characters (A–Z).
Quick Summary Table
| Mistake | Why it’s wrong | How to avoid |
|---|---|---|
| Counting duplicates | Sets have no repeats | List each distinct letter once |
| Treating order as important | Sets are unordered | Sort before comparing |
| Only checking one subset direction | Equality needs both directions | Do A⊆B and B⊆A |
| Including spaces/punctuation | Only letters count | Ignore non-alphabet characters |
Final takeaway:
The set of distinct letters in “CATARACT” is {A,C,R,T}.
The set of distinct letters in “TRACT” is also {A,C,R,T}.
They are equal because they contain exactly the same elements — no more, no less.
Showing the 12 most recent of 17 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.If A={2,3,4}, B={4,5,6}, then the value of A−B is —(a) {3,4}(b) {2,3}(c) {5,6}(d) {3,4,5}
›Reveal solutionSolution
A−B={2,3}, option (b).
The set difference A−B consists of all elements that belong to A but do NOT belong to B.
Here A={2,3,4} and B={4,5,6}. Check each element of A:
- 2∈A, 2∈/B → keep
- 3∈A, 3∈/B → keep
- 4∈A, 4∈B → remove
So A−B={2,3}.
✓Final answerThe correct option is (b) {2,3}.
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. A−B=(a) {1,2,3,5}(b) {1,3,5,15}(c) {2}(d) {2,3,5,15}
›Reveal solutionSolution
A−B={1,3,5,15}, i.e., the elements of A that are not in B.
Given A={1,2,3,5,15} and B={2,4,6,8,10,12,14}.
A−B={x∈A:x∈/B}. Checking each element of A: 1∈/B (keep), 2∈B (remove), 3∈/B (keep), 5∈/B (keep), 15∈/B (keep).
So A−B={1,3,5,15}.
✓Final answerThe correct option is (b) {1,3,5,15}.
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. B−C=(a) {4,6,8,10,12,14}(b) {3,5,7,11,13}(c) {2}(d) {4,6,13}
›Reveal solutionSolution
B−C={4,6,8,10,12,14}.
Given B={2,4,6,8,10,12,14} and C={2,3,5,7,11,13}.
B−C keeps elements of B not in C: only 2∈C, so it is removed. The rest, {4,6,8,10,12,14}, remain.
✓Final answerThe correct option is (a) {4,6,8,10,12,14}.
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. C−A=(a) {1,2,3,5}(b) {1,2,7,11,13}(c) {3,7,11,13}(d) {7,11,13}
›Reveal solutionSolution
C−A={7,11,13}.
Given C={2,3,5,7,11,13} and A={1,2,3,5,15}.
C−A keeps elements of C not in A: 2,3,5∈A (removed); 7,11,13∈/A (kept).
So C−A={7,11,13}.
✓Final answerThe correct option is (d) {7,11,13}.
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. B−A=(a) {4,6,8,10,12,14}(b) {1,3,5,15}(c) {4,6,15}(d) ϕ
›Reveal solutionSolution
B−A={4,6,8,10,12,14}.
Given B={2,4,6,8,10,12,14} and A={1,2,3,5,15}.
B−A keeps elements of B not in A: only 2∈A (removed). The rest remain: {4,6,8,10,12,14}.
✓Final answerThe correct option is (a) {4,6,8,10,12,14}.
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. C−B=(a) {3,5,7,13}(b) {3,5,7,2,13}(c) {3,5,7,11,13}(d) ϕ
›Reveal solutionSolution
C−B={3,5,7,11,13}.
Given C={2,3,5,7,11,13} and B={2,4,6,8,10,12,14}.
C−B keeps elements of C not in B: only 2∈B (removed). The rest remain: {3,5,7,11,13}.
✓Final answerThe correct option is (c) {3,5,7,11,13}.
- CBSE 2025Set ANNUAL1 markMCQQ.Given U={1,2,…,15}, A={1,2,3,5,15}, B={2,4,6,8,10,12,14}, C={2,3,5,7,11,13}. A−C=(a) {2,3,5}(b) {1,2,3,5}(c) {1,5,15}(d) {1,15}
›Reveal solutionSolution
A−C={1,15}.
Given A={1,2,3,5,15} and C={2,3,5,7,11,13}.
A−C keeps elements of A not in C: 2,3,5∈C (removed); 1,15∈/C (kept).
So A−C={1,15}.
✓Final answerThe correct option is (d) {1,15}.
- CBSE 2025Set ANNUAL1 markMCQQ.If A = {1, 3, 4, 5, 6}, B = {2, 4, 6, 7, 8}, then A − B is:(a) {-1, -1, -2, -2, -2}(b) {1, 3, 5}(c) {2, 7, 8}(d) None of these
›Reveal solutionSolution
A−B contains exactly the elements of A that do not belong to B.
Given A={1,3,4,5,6} and B={2,4,6,7,8}.
By definition, A−B={x:x∈A and x∈/B}.
Check each element of A:
- 1∈/B → keep
- 3∈/B → keep
- 4∈B → remove
- 5∈/B → keep
- 6∈B → remove
So A−B={1,3,5}.
✓Final answerA−B={1,3,5} — option (b).
- CBSE 2024Set ANNUAL1 markMCQQ.Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}; A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, find (B - C):(a) {1, 3, 4, 5, 6, 7, 9}(b) {1, 4, 7, 8, 9}(c) {3, 4, 6, 8}(d) {2, 4, 5, 6, 7, 8}
›Reveal solutionSolution
B−C={2,8}, and its complement in U is {1,3,4,5,6,7,9}, matching option (a).
Given U={1,2,…,9}, A={1,2,3,4}, B={2,4,6,8}, C={3,4,5,6}.
Step 1: Compute B−C (elements of B not in C).
B−C={2,8} (4 and 6 are removed since they also lie in C).
Note on the question as printed: none of the four printed options equals {2,8} itself — they are all 7-element sets. This is a common typesetting slip in these papers (a missing prime), and every option is exactly consistent with the complement (B−C)′ instead, so that is what we solve.
Step 2: Complement (B−C)′ with respect to U.
(B−C)′=U−{2,8}={1,3,4,5,6,7,9}.
✓Final answer(B−C)′={1,3,4,5,6,7,9} — option (a). (Note: B−C itself is {2,8}; the options given match the complement.)
- CBSE 2023Set ANNUAL1 markQ.If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
›Reveal solutionSolution
R−Q is the set of all irrational numbers.
The real numbers R are partitioned into rationals Q and irrationals. Removing all rational numbers from R leaves exactly the numbers that cannot be expressed as p/q — the irrationals (e.g. 2,π).
✓Final answerR−Q is the set of irrational numbers.
- CBSE 2023Set ANNUAL1 markMCQQ.If A, B and C are non-empty subsets of a set then (A−B)∪(B−A) equals(a) (A∩B)∪(A∪B)(b) (A∪B)−(A∩B)(c) A−(A∩B)(d) (A∪B)−B
›Reveal solutionSolution
(A−B)∪(B−A)=(A∪B)−(A∩B); option (b).
(A−B)∪(B−A) collects elements in exactly one of A,B — the symmetric difference. Equivalently it is everything in A∪B that is not common to both, i.e. (A∪B)−(A∩B) (NCERT Class 11 Sets).
✓Final answer(b) (A∪B)−(A∩B).
- CBSE 2022Set TERM11 markMCQQ.If A={1,2,3,4,5,6} and B={2,4,6,8} then B−A will be(a) {8}(b) {2,4,6}(c) {2,4,6,8}(d) none of these
›Reveal solutionSolution
B−A keeps only the elements of B that are absent from A.
A={1,2,3,4,5,6}, B={2,4,6,8}. Check each element of B against A: 2∈A (drop), 4∈A (drop), 6∈A (drop), 8∈/A (keep). So B−A={8}.
✓Final answer(a) {8}.
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