RD Sharma Solutions for Class 11 Chapter 17 – Combinations are designed to support students in mastering key concepts and performing well in their final examinations. In earlier chapters, students learned about arranging a certain number of objects either partially or completely. In contrast, this chapter focuses on selections, where the order does not matter, and such selections are referred to as combinations.
These solutions are especially helpful for self-learners who prefer to work independently. With clearly explained steps and accurate answers, these resources serve as a reliable guide for tackling each exercise problem confidently.
RD Sharma Solutions Class 11 Maths Chapter 17 – Exercise wise
Chapter 17 includes three well-structured exercises, each addressing different aspects of combinations. The solutions provided here have been carefully curated by academic professionals, following the latest CBSE exam pattern and curriculum requirements.
All solutions are available for free download in PDF format, enabling students to access them anytime and anywhere for better preparation and practice.
- Definition of Combinations
- Important Properties of C(n, r)
- Application-Based Questions on Combinations
- Mixed Problems Involving Both Permutations and Combinations
RD Sharma Solutions for Class 11 Maths Combinations - Question with solutions
Basic Conceptual Questions
- Find the value of ⁵C₂.
⁵C₂ = 5! / [2!(5−2)!] = 120 / (2×6) = 10 - Evaluate ⁷C₄.
⁷C₄ = 7! / (4!×3!) = 5040 / (24×6) = 35 - If ⁸C₃ = ⁸C₅, verify the result.
True, because nCr = nC(n−r) - Find the value of ¹⁰C₇.
¹⁰C₇ = 10! / (7!×3!) = 120 - If nC₃ = 120, find n.
n(n−1)(n−2)/6 = 120 ⇒ Try n = 10 ⇒ n = 10
Intermediate Application Questions
- How many ways can you choose 3 people from 8?
⁸C₃ = 56 - From a group of 12 students, how many ways can you form a team of 5?
¹²C₅ = 792 - Form a 4-member committee with at least 1 woman from 6 men and 4 women.
Total = ¹⁰C₄ = 210, All men = ⁶C₄ = 15 ⇒ Required = 195 - Select 6 books from 10.
¹⁰C₆ = 210 - If ⁿC₄ = ⁿC₆, find n.
n = 4 + 6 = 10
Word Problems with Combinations
- Choose 6 out of 10 questions.
¹⁰C₆ = 210 - Committee with 2 men and 2 women from 7 men and 5 women.
⁷C₂ × ⁵C₂ = 21 × 10 = 210 - Select 4 balls from 10 different balls.
¹⁰C₄ = 210 - Team of 2 boys and 3 girls from 5 boys and 6 girls.
⁵C₂ × ⁶C₃ = 10 × 20 = 200 - Committee of 3 from 4 boys and 3 girls with at least 1 boy.
Total = ⁷C₃ = 35, All girls = ³C₃ = 1 ⇒ Required = 34
Property-Based Questions
- Show nC₀ + nC₁ + … + nCn = 2ⁿ for n = 4.
1 + 4 + 6 + 4 + 1 = 16 = 2⁴ - Prove nCn = 1.
nCn = n! / (n!(n−n)!) = 1 - Find ⁹C₆ using identity.
⁹C₆ = ⁹C₃ = 84 - If nC₂ = 66, find n.
n(n−1)/2 = 66 ⇒ n = 12 - Evaluate: ¹⁰C₀ + ¹⁰C₁ + … + ¹⁰C₁₀
= 2¹⁰ = 1024
Higher Order Thinking (HOTS)
- 4-digit numbers from 5 digits, no repetition.
⁵C₄ × 4! = 5 × 24 = 120 - Select 5 balls with at least 3 red from 6 red, 4 green.
Total = 186 - Number of diagonals in a 10-sided polygon.
ⁿC₂ − n = 45 − 10 = 35 - Ways to distribute 5 prizes to 10 students (1 each).
¹⁰C₅ = 252 - Choose 6 balls including 2 red from 4 red, 8 white.
⁴C₂ × ⁸C₄ = 6 × 70 = 420
Miscellaneous Questions
- Find r if ⁷Cr = ⁷C(r+2).
Try r = 2 → ⁷C₂ = 21 = ⁷C₅ ⇒ r = 2 - Evaluate (n+1)C₃ − nC₂.
Answer = nC₃ - Number of triangles from 8 non-collinear points.
⁸C₃ = 56 - 3-letter combinations from "TRIANGLE".
⁸C₃ = 56 - How many committees of 3 from 5 people?
⁵C₃ = 10
Frequently Asked Questions
Combinations are a critical concept in Class 11 that form the base for Probability, Binomial Theorem, and Set Theory.
RD Sharma provides step-by-step, conceptual clarity which makes it ideal for board preparation.
The questions are graded in difficulty, enabling students to go from beginner to advanced level gradually.
Practicing this chapter helps students gain confidence in solving selection-based problems effectively.
Ideally, students should solve at least 30–40 varied questions from this chapter, covering all types like basic, applied, and HOTS.
RD Sharma’s exercises, along with examples and previous year questions, provide a solid base.
For competitive exams like JEE, practicing combination-based problems involving constraints (e.g., “at least one girl”, “exactly two red balls”) is highly recommended.
The chapter includes a mix of:
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Basic formula-based questions
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Application in word problems (e.g., selection of students, committees)
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Identity-based questions like nCr = nC(n−r)
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Properties and sum of combinations: nC₀ + nC₁ + … + nCn = 2ⁿ
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Higher-order thinking (HOTS) questions
These types ensure students build a strong foundational understanding for competitive exams like JEE or Olympiads.
The key difference lies in the importance of order:
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Permutation considers the order of selection (e.g., seating arrangements).
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Combination ignores the order – it's all about selection (e.g., selecting a committee).
Mathematically:
nPr = n! / (n − r)!
vs
nCr = n! / [r!(n − r)!]
RD Sharma clearly explains both with solved examples to strengthen conceptual clarity.