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RD Sharma Solutions for Class 11 Chapter 17 Combinations | Free PDF Download
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Updated on 30 May 2025, 10:53 IST
RD Sharma Solutions for Class 11 Mathematics Chapter “Combinations” are available here, prepared by expert teachers according to the latest CBSE Board syllabus and NCERT guidelines. This chapter introduces the fundamental concept of combinations, which is essential for understanding how to select objects from a group, regardless of their arrangement. On this page, students will find a wide variety of combination problems with detailed, step-by-step solutions to help master this important topic.
These RD Sharma Solutions for Class 11 Combinations solutions assist students in learning how to calculate combinations of distinct objects, understand and use combination formulas, and apply properties of combinations in real-life scenarios. Mastering combinations is crucial for excelling in board exams, competitive exams, and further studies in mathematics and related fields.
RD Sharma Solutions for Class 11 Chapter 17 Combinations
In RD Sharma Solutions for Class 11 Chapter Combinations, students will gain a clear understanding of the core concept of combinations and how it differs from permutations. The chapter covers how to calculate combinations of n objects taken r at a time, properties of combinations, and practical problems involving combinations. Students also learn to apply combination formulas to solve problems involving selection of items, groups, and teams.
Important topics include the derivation and properties of C(n,r), mixed problems on permutations and combinations, and detailed, step-by-step solutions to all textbook exercises for thorough practice. This chapter builds a strong foundation in combinatorial mathematics, enabling students to confidently analyze and solve complex combination problems essential for both board exams and competitive tests.
Download RD Sharma Solutions for Combinations PDF Free
Download comprehensive RD Sharma Solutons for Class 11 Maths solutions for the chapter on combinations, including solved examples, extra practice questions, and detailed explanations. Access the free PDF to enhance your understanding of combinations, improve your accuracy, and boost your exam preparation with trusted, exam-oriented solutions.
RD Sharma Solutions for Combination: Solved Examples
Question 1: Evaluate
(i) 14C3
Solution:
Using the formula: nCr = n!/(r!(n-r)!)
14C3 = 14!/(3!(14-3)!) = 14!/(3! × 11!)
= (14 × 13 × 12 × 11!)/(3! × 11!)
= (14 × 13 × 12)/(3 × 2 × 1)
= (14 × 13 × 12)/6 = 2184/6 = 364
(ii) 35C35
Solution:
35C35 = 35!/(35!(35-35)!) = 35!/(35! × 0!) = 1/1 = 1
(iii) n+1Cn
Solution:
n+1Cn = (n+1)!/(n!(n+1-n)!) = (n+1)!/(n! × 1!) = (n+1)/1 = n+1
Question 2: Solve for n
nC12 = nC5
Solution:
Using the property: If nCr = nCs, then either r = s or r = n - s
Here, 12 ≠ 5, so 12 = n - 5
n = 12 + 5 = 17
Verification: 17C12 = 17C5 ✓
Question 3: Find r
15C3r = 15Cr+3
Solution:
Using the property: If nCx = nCy, then either x = y or x = n - y
Case 1: 3r = r + 3
2r = 3, so r = 3/2 (not an integer, invalid)
Case 2: 3r = 15 - (r + 3) = 12 - r
4r = 12, so r = 3
Verification: 15C9 = 15C6 ✓
Question 4: Prove
nCr + nCr-1 = n+1Cr
Solution:
LHS = nCr + nCr-1
= n!/(r!(n-r)!) + n!/((r-1)!(n-r+1)!)
= n!/((r-1)!(n-r)!) × [1/r + 1/(n-r+1)]
= n!/((r-1)!(n-r)!) × [(n-r+1+r)/(r(n-r+1))]
= n!/((r-1)!(n-r)!) × (n+1)/(r(n-r+1))
= (n+1)!/(r!(n+1-r)!) = n+1Cr = RHS
Hence proved.
Question 5
If 20Cr = 20Cr-10, find 18Cr
Solution:
Using the property: If nCx = nCy, then either x = y or x = n - y
Case 1: r = r - 10 (impossible)
Case 2: r = 20 - (r - 10) = 30 - r
2r = 30, so r = 15
Therefore, 18C15 = 18C3 = (18 × 17 × 16)/(3 × 2 × 1) = 4896/6 = 816
Question 6
How many ways can 5 boys and 3 girls be selected from 25 boys and 10 girls for a boat party?
Solution:
Number of ways to select 5 boys from 25 = 25C5
Number of ways to select 3 girls from 10 = 10C3
Total ways = 25C5 × 10C3
25C5 = (25 × 24 × 23 × 22 × 21)/(5 × 4 × 3 × 2 × 1) = 53,130
10C3 = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120
Total ways = 53,130 × 120 = 6,375,600
Question 7
Calculate the number of diagonals in a 12-sided polygon.
Solution:
Total number of line segments joining any 2 vertices = 12C2
Number of sides = 12
Number of diagonals = 12C2 - 12
12C2 = (12 × 11)/2 = 66
Number of diagonals = 66 - 12 = 54
Question 8
Write the simplified form of: Σr=0m n+rCr
Solution:
This is a well-known identity in combinatorics:
Σr=0m n+rCr = n+m+1Cm
Question 9
From a group of 15 cricket players, how many ways can a team of 11 players be chosen?
Solution:
Number of ways = 15C11
15C11 = 15C4 = (15 × 14 × 13 × 12)/(4 × 3 × 2 × 1)
= 32,760/24 = 1,365
Question 10
If 18Cx = 18Cx+2, find x
Solution:
Using the property: If nCa = nCb, then either a = b or a = n - b
Case 1: x = x + 2 (impossible)
Case 2: x = 18 - (x + 2) = 16 - x
2x = 16, so x = 8
Verification: 18C8 = 18C10 ✓
Question 11
How many parallelograms can be formed by 4 parallel lines intersecting 3 other parallel lines?
Solution:
To form a parallelogram, we need to select 2 lines from the first set of 4 parallel lines and 2 lines from the second set of 3 parallel lines.
Number of ways to select 2 from 4 = 4C2 = 6
Number of ways to select 2 from 3 = 3C2 = 3
Total parallelograms = 6 × 3 = 18
Question 12
Find the number of ways to distribute 12 distinct books into 3 groups of 4 books each.
Solution:
This is a problem of dividing objects into groups.
Number of ways = 12C4 × 8C4 × 4C4 / 3!
Note: We divide by 3! because the groups are indistinguishable.
12C4 = 495, 8C4 = 70, 4C4 = 1
Number of ways = (495 × 70 × 1)/6 = 34,650/6 = 5,775
Question 13
Solve: (8Cr - 7C2)/7C3 = 1
Solution:
8Cr - 7C2 = 7C3
7C2 = 21, 7C3 = 35
8Cr = 35 + 21 = 56
We need to find r such that 8Cr = 56
Testing values: 8C3 = 56 ✓
Therefore, r = 3 (or r = 5, since 8C5 = 8C3)
Question 14
How many 5-card hands can be drawn from a standard deck of 52 cards?
Solution:
Number of ways = 52C5
52C5 = (52 × 51 × 50 × 49 × 48)/(5 × 4 × 3 × 2 × 1)
= 311,875,200/120 = 2,598,960
Question 15
If 35Cn+7 = 35C4n-2, find n
Solution:
Using the property: If aCx = aCy, then either x = y or x = a - y
Case 1: n + 7 = 4n - 2
9 = 3n, so n = 3
Case 2: n + 7 = 35 - (4n - 2) = 37 - 4n
5n = 30, so n = 6
We need to check which values are valid (0 ≤ n+7 ≤ 35 and 0 ≤ 4n-2 ≤ 35)
For n = 3: n+7 = 10, 4n-2 = 10 ✓
For n = 6: n+7 = 13, 4n-2 = 22 ✓
Both solutions are valid: n = 3 or n = 6
Question 16
Determine the number of ways to select 4 vowels and 5 consonants from 7 vowels and 9 consonants.
Solution:
Number of ways to select 4 vowels from 7 = 7C4
Number of ways to select 5 consonants from 9 = 9C5
Total ways = 7C4 × 9C5
7C4 = 7C3 = 35
9C5 = 9C4 = 126
Total ways = 35 × 126 = 4,410
Question 17
A committee of 7 must include at least 3 women. How many ways can this be done from 10 men and 8 women?
Solution:
Total ways = Ways with exactly 3 women + Ways with exactly 4 women + Ways with exactly 5 women + Ways with exactly 6 women + Ways with exactly 7 women
Case 1: 3 women, 4 men = 8C3 × 10C4 = 56 × 210 = 11,760
Case 2: 4 women, 3 men = 8C4 × 10C3 = 70 × 120 = 8,400
Case 3: 5 women, 2 men = 8C5 × 10C2 = 56 × 45 = 2,520
Case 4: 6 women, 1 man = 8C6 × 10C1 = 28 × 10 = 280
Case 5: 7 women, 0 men = 8C7 × 10C0 = 8 × 1 = 8
Total = 11,760 + 8,400 + 2,520 + 280 + 8 = 22,968
Question 18
Find the maximum number of intersection points formed by 10 straight lines in a plane.
Solution:
Maximum intersection points occur when no two lines are parallel and no three lines meet at a point.
Each pair of lines intersects at exactly one point.
Number of intersection points = 10C2 = (10 × 9)/2 = 45
Question 19
How many triangles can be formed by joining 18 non-collinear points?
Solution:
To form a triangle, we need to select any 3 points from 18 non-collinear points.
Number of triangles = 18C3
18C3 = (18 × 17 × 16)/(3 × 2 × 1) = 4,896/6 = 816
Question 20: Mixed Problem
How many 4-digit numbers can be formed using digits 1, 2, 3, 4, 5 if repetitions are allowed, and how many are even?
Solution:
Part 1: Total 4-digit numbers with repetition allowed
Each position can be filled in 5 ways (any of 1, 2, 3, 4, 5)
Total numbers = 5 × 5 × 5 × 5 = 625
Part 2: Even numbers
For a number to be even, the last digit must be even.
Even digits available: 2, 4 (2 choices)
First three positions: 5 choices each
Even numbers = 5 × 5 × 5 × 2 = 250
Answer: Total 4-digit numbers = 625, Even numbers = 250
FAQs: RD Sharma Solutions for Class 11 Chapter 17 Combinations
What is a combination in mathematics, and how is it different from a permutation?
A combination is the selection of items from a group, where the order does not matter. In contrast, a permutation considers the arrangement or order of the selected items.
How many exercises are there in RD Sharma Class 11 Chapter 17 Combinations?
There are three exercises in Chapter 17 – Combinations, each covering different aspects like properties of combinations, practical problems, and mixed problems with permutations.
Why should I use RD Sharma Solutions for Class 11 Combinations?
RD Sharma Solutions provide detailed, step-by-step answers to all textbook questions, making it easier to understand concepts, clarify doubts, and prepare effectively for exams. They are created by subject experts as per the latest CBSE guidelines.
Is the PDF of RD Sharma Solutions for Combinations available for free?
Yes at Infinity Learn, students can download the RD Sharma Class 11 Combinations Solutions PDF for free from leading educational websites.
What are the key topics covered in the Combinations chapter of RD Sharma?
Key topics include the definition and formula of combinations, properties of C(n,r), practical problems, and mixed questions involving both permutations and combinations.
How does practising RD Sharma Solutions help in board exam preparation?
Practising these solutions helps students gain a deeper understanding of concepts, improve speed and accuracy, and become familiar with the types of questions frequently asked in board exams.