Solution: Using nCr = nCn−r: if nC8 = nC6, then either 8 = 6 (impossible) or 8 + 6 = n ∴ n = 14
📋 Assertion-Reason Questions (AR1–AR5)
📌 Instructions
In each question below, Statement I is the Assertion (A) and Statement II is the Reason (R). Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
AR Question 1Sets · Subsets
Assertion (A): The total number of subsets of a set with n elements is 2ⁿ.
Reason (R): For each element of the set, there are exactly 2 choices — either include it in a subset or exclude it — and these choices are independent for every element.
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
✓ Correct Answer: (A)
Explanation: A is true: a set with n elements has exactly 2ⁿ subsets. ✓ R is true and correctly explains A: each of the n elements is independently either included or excluded, giving 2 × 2 × … × 2 (n times) = 2ⁿ combinations.
AR Question 2Relations · Cartesian Product
Assertion (A): If A = {1, 2, 3} and B = {a, b}, then n(A × B) = 6.
Reason (R): The Cartesian product A × B is defined as the set of all ordered pairs (x, y) such that x ∈ B and y ∈ A.
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
✓ Correct Answer: (C)
Explanation: A is true: n(A × B) = n(A) × n(B) = 3 × 2 = 6. ✓ R is false: The definition in R has the pair reversed. The correct definition is A × B = {(x, y) : x ∈ A and y ∈ B}, not x ∈ B and y ∈ A. Since A is true but R is false → Option (C).
AR Question 3Arithmetic Progression
Assertion (A): The sum of the first n natural numbers is n(n + 1)/2.
Reason (R): The natural numbers 1, 2, 3, …, n form an AP with first term a = 1 and common difference d = 1, and applying Sn = n/2 × [2a + (n−1)d] gives n(n+1)/2.
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
✓ Correct Answer: (A)
Explanation: A is true: 1 + 2 + … + n = n(n+1)/2. ✓ R is true and correctly explains A: with a = 1, d = 1 → Sn = n/2 × [2(1) + (n−1)(1)] = n/2 × (n+1) = n(n+1)/2. ✓
AR Question 4Geometric Progression
Assertion (A): The sum of the infinite GP 1 + 1/3 + 1/9 + 1/27 + … is 3/2.
Reason (R): The sum of an infinite GP exists only when |r| > 1, and is given by S∞ = a / (1 − r).
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
✓ Correct Answer: (C)
Explanation: A is true: a = 1, r = 1/3. Since |r| < 1, S∞ = 1 / (1 − 1/3) = 1 / (2/3) = 3/2. ✓ R is false: The condition in R is wrong. An infinite GP converges (and its sum exists) when |r| < 1, not |r| > 1. The formula S∞ = a/(1−r) is correct, but the condition stated is incorrect. → Option (C).
AR Question 5Permutations & Combinations
Assertion (A):nCr = nCn−r for all valid values of n and r.
Reason (R): Selecting r objects from n to include in a group is equivalent to selecting the remaining (n − r) objects to exclude — both describe the same division of n objects into two groups.
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
✓ Correct Answer: (A)
Explanation: A is true: nCr = n!/[r!(n−r)!] = n!/[(n−r)!r!] = nCn−r. ✓ R is true and correctly explains A: choosing which r objects to include automatically decides which (n−r) objects are excluded. Both perspectives count the same set of selections, so nCr = nCn−r. ✓
📐 All Unit 2 formulas — Sets, AP, GP, P&C — in one printable revision PDF
Short Answer Questions with Step-by-Step Solutions
13 questions across all 4 sections · 2-mark & 3-mark level
📦 Section A — Sets & Relations
Question 1Sets · Operations
If A = {x : x is an even natural number ≤ 10} and B = {x : x is a prime natural number ≤ 10}, find A ∪ B, A ∩ B and A – B.
Solution:
A = {2, 4, 6, 8, 10} B = {2, 3, 5, 7}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 10}
A ∩ B = {2} (only element common to both)
A – B = {4, 6, 8, 10} (elements in A not in B)
A ∪ B = {2,3,4,5,6,7,8,10} | A ∩ B = {2} | A – B = {4,6,8,10}
Question 2Sets · De Morgan's Law
Let U = {1, 2, …, 10}, A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7}. Verify: (A ∪ B)' = A' ∩ B'.
Solution:
A ∪ B = {1,2,3,5,7,9} ∴ (A ∪ B)' = {4,6,8,10} ...(LHS)
A' = {2,4,6,8,10} B' = {1,4,6,8,9,10}
A' ∩ B' = {4,6,8,10} ...(RHS)
LHS = RHS = {4,6,8,10} ✓ De Morgan's Law Verified
Question 3Sets · Intervals
Write the following sets as intervals and state whether each interval is open, closed, semi-open or semi-closed: (a) {x ∈ ℝ : −3 < x ≤ 5} (b) {x ∈ ℝ : 0 ≤ x ≤ 4} (c) {x ∈ ℝ : −1 < x < 2}
Solution:
(a) {x : −3 < x ≤ 5} = (−3, 5] → Semi-closed interval. Left endpoint excluded, right included.
(b) {x : 0 ≤ x ≤ 4} = [0, 4] → Closed interval. Both endpoints included.
(c) {x : −1 < x < 2} = (−1, 2) → Open interval. Both endpoints excluded.
In a group of 75 students, 45 study Mathematics, 30 study Biology and 15 study both. Using a Venn diagram, find: (i) the number who study at least one subject, (ii) only Mathematics, (iii) only Biology, and (iv) neither subject.
Solution:
Let M = Mathematics, B = Biology. n(M) = 45, n(B) = 30, n(M ∩ B) = 15, Total = 75
(i) n(M ∪ B) = 45 + 30 − 15 = 60
(ii) Only Mathematics = 45 − 15 = 30
(iii) Only Biology = 30 − 15 = 15
(iv) Neither = 75 − 60 = 15
(i) 60 | (ii) Only Maths: 30 | (iii) Only Biology: 15 | (iv) Neither: 15
Question 5Sets · Subsets & Power Set
If A = {p, q, r}, find: (i) all subsets of A, (ii) the power set P(A), (iii) the number of proper subsets of A, and (iv) how many subsets of A have exactly 2 elements.
(a) S∞ = 4.5 (b) AM = 6.5, GM = 6; AM ≥ GM verified ✓
🔢 Section C — Permutations & Combinations
Question 10FPC · Fundamental Principles
A person wants to travel from city A to city C via city B. There are 3 routes from A to B and 4 routes from B to C. In how many ways can the person make the trip (i) from A to C and (ii) from A to C and back, but returning via a different route from C to B?
(ii) Return trip C→B has 4 routes; but must be a different one → 3 choices; then B→A: 3 choices
Return: 3 × 3 = 9 ways Total: 12 × 9 = 108 ways
(i) 12 ways (ii) 108 ways
Question 11Permutations
How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 if (i) repetition is allowed and (ii) repetition is not allowed?
Solution:
(i) With repetition: Each of the 3 positions can be filled by any of 5 digits. Total = 5 × 5 × 5 = 125
(ii) Without repetition:5P3 = 5!/(5−3)! = 5!/2! = 5×4×3 = 60
(i) 125 (ii) 60
Question 12Combinations
A committee of 5 is to be formed from 6 men and 4 women. In how many ways can this be done so that the committee has (i) exactly 2 women and (ii) at least 2 women?
Solution:
(i) Exactly 2 women: Choose 2 women from 4: 4C2 = 6 Choose 3 men from 6: 6C3 = 20 Total = 6 × 20 = 120
(ii) At least 2 women (2W+3M, 3W+2M, 4W+1M): 2W+3M: 4C2 × 6C3 = 6×20 = 120 3W+2M: 4C3 × 6C2 = 4×15 = 60 4W+1M: 4C4 × 6C1 = 1×6 = 6 Total = 120 + 60 + 6 = 186
(i) 120 (ii) 186
Question 13Combinations · Key Result
Prove that nCr + nCr−1 = n+1Cr. Verify for n = 5, r = 3.
4 real-world application based questions — Sets, AP, GP and P&C
Case Study 1Sets · Venn Diagram
School Sports Survey
A school conducted a sports preference survey among its 200 Class 11 students. The results showed that 60% of students like Cricket, 45% like Football, and 20% like both Cricket and Football. Some students like neither sport. Using this information, answer the following questions. Let C denote the set of students who like Cricket and F the set of students who like Football.
(a) Find the number of students who like Cricket, Football, and both.
Solution:
n(C) = 60% of 200 = 120
n(F) = 45% of 200 = 90
n(C ∩ F) = 20% of 200 = 40
Cricket: 120 students | Football: 90 students | Both: 40 students
(b) How many students like at least one sport? What percentage is this?
Solution:
n(C ∪ F) = 120 + 90 − 40 = 170
Percentage = (170/200) × 100 = 85%
170 students (85%) like at least one sport
(c) How many students like only Cricket (not Football) and how many like only Football (not Cricket)?
Solution:
Only Cricket = 120 − 40 = 80
Only Football = 90 − 40 = 50
Only Cricket: 80 students | Only Football: 50 students
(d)(i) How many students like neither Cricket nor Football? Express as a percentage.
Solution:
Neither = 200 − 170 = 30
Percentage = (30/200) × 100 = 15%
30 students (15%) like neither sport
(d)(ii) OR: Verify using De Morgan's Law that (C ∪ F)' = C' ∩ F' by finding n(C') and n(F').
Solution:
n(C') = 200 − 120 = 80 (students who do NOT like Cricket)
n(F') = 200 − 90 = 110 (students who do NOT like Football)
n(C' ∩ F') = students in neither = 30 ...(RHS)
n(C ∪ F)' = 200 − 170 = 30 ...(LHS)
LHS = RHS = 30 ✓ De Morgan's Law: (C ∪ F)' = C' ∩ F' verified
Case Study 2Arithmetic Progression
Salary Growth in a Company
Rajan joins a company with a starting monthly salary of ₹25,000. The company offers a fixed annual increment of ₹1,500. His monthly salary each year forms an Arithmetic Progression. Let n represent the year number (n = 1 for the first year).
(a) Write the AP and identify a and d.
Solution:
AP: 25000, 26500, 28000, 29500, ...
First term a = ₹25,000 Common difference d = ₹1,500
(b) What will be Rajan's monthly salary in the 8th year?
Solution:
a8 = 25000 + 7×1500 = ₹35,500
Monthly salary in 8th year = ₹35,500
(c) Find the total salary Rajan earns in the first 10 years.
(d)(i) In which year will Rajan's monthly salary first exceed ₹40,000?
Solution:
25000 + (n−1)×1500 > 40000 → (n−1) > 10 → n > 11
Salary first exceeds ₹40,000 in the 12th year
(d)(ii) OR: What is the arithmetic mean of Rajan's salaries in the 5th and 15th years?
Solution:
a5 = 25000 + 4×1500 = ₹31,000
a15 = 25000 + 14×1500 = ₹46,000
AM = (31000 + 46000)/2 = ₹38,500
AM of salaries in 5th and 15th years = ₹38,500 Note: This also equals a10 = 25000 + 9×1500 = ₹38,500 — the middle term property of AP.
Case Study 3Geometric Progression · Application
Virus Spread Model
During a disease outbreak, health officials observe that the number of new cases reported each day follows a Geometric Progression. On Day 1, there were 5 new cases. Each subsequent day the number of new cases triples. Let an denote the number of new cases on Day n.
(a) Identify a and r for the GP and write the first five terms.
Solution:
First term a = 5, Common ratio r = 3
a = 5, r = 3. First five terms: 5, 15, 45, 135, 405
(b) How many new cases are reported on Day 7?
Solution:
a7 = ar⁶ = 5 × 3⁶ = 5 × 729 = 3645
New cases on Day 7 = 3,645
(c) Find the total number of new cases reported over the first 6 days.
(d)(i) On which day do new cases first exceed 10,000?
Solution:
5 × 3n−1 > 10000 → 3n−1 > 2000
3⁶ = 729, 3⁷ = 2187 > 2000 → n − 1 = 7 → n = 8
New cases first exceed 10,000 on Day 8
(d)(ii) OR: Find the GM of cases on Day 1 and Day 3 and verify it equals cases on Day 2.
Solution:
a1 = 5, a3 = 45 GM = √(5 × 45) = √225 = 15
a2 = 15 ✓ (GM of two terms of a GP equals the middle term)
GM = 15 = a2 ✓ Verified
Case Study 4Permutations & Combinations
School Annual Day Planning
A school is organising its Annual Day. The event committee has 8 students — 5 from Class 11 and 3 from Class 12. They need to select teams and make various arrangements for different tasks as described below.
(a) A team of 3 students is to be selected from all 8. In how many ways can this be done?
Solution:
Order doesn't matter → Combinations
8C3 = (8×7×6)/(3×2×1) = 56
Number of ways = 56
(b) 4 students are to be arranged in a row for the welcome banner photo (selected from all 8). In how many ways can this be done?
Solution:
Order matters → Permutations
8P4 = 8×7×6×5 = 1680
Number of arrangements = 1,680
(c) In how many ways can a sub-team of exactly 2 Class 11 and 1 Class 12 student be selected?
Solution:
5C2 × 3C1 = 10 × 3 = 30
Number of ways = 30
(d)(i) The MC must deliver 3 different speeches, one opening, one during, and one closing. In how many ways can the MC arrange 3 speeches chosen from 7 prepared ones?
Solution:
Order matters → Permutations
7P3 = 7×6×5 = 210
Number of arrangements = 210
(d)(ii) OR: The event has 4 cultural acts and 3 dance performances. The MC must pick 1 cultural act and 1 dance performance to open the show. In how many ways can this be done?
Solution:
By Fundamental Principle of Multiplication:
Ways = 4 × 3 = 12
Number of ways = 12
📐 Ace Unit 2 — Get the Class 11 Formula Deck
Every formula from all 8 units of Class 11 Applied Maths — organised, concise, ready to print.
What's inside the Formula Deck?
✅ All formulas for all 8 units — organised topic-wise
✅ Sets laws, De Morgan's, AP & GP formulas, P&C formulas in full
✅ Important theory points & key results from MCQ perspective
✅ Concise & comprehensive as per CBSE Syllabus Code 241 (2025–26)
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