Class 11 Mathematics Notes Chapter 7 (Chapter 7) – Examplar Problems (English) Book

Alright class, let's dive into Chapter 7: Permutations and Combinations from your NCERT Exemplar. This is a crucial chapter, not just for your Class 11 exams, but also because concepts from here frequently appear in various government recruitment exams. Pay close attention as we break down the core ideas.

Chapter 7: Permutations and Combinations - Detailed Notes for Government Exam Preparation

Core Idea: This chapter deals with methods of counting the number of ways events can occur, either by arranging items (Permutations) or selecting items (Combinations).

1. Fundamental Principle of Counting (FPC)

This is the bedrock of the entire chapter.

• Multiplication Principle: If an event 'A' can occur in 'm' different ways, and following it, another independent event 'B' can occur in 'n' different ways, then the total number of ways both events can occur in the stated order is m × n. This extends to any finite number of events.
  • Example: Choosing 1 shirt out of 3 and 1 pair of pants out of 2. Total outfits = 3 × 2 = 6.
• Addition Principle: If an event 'A' can occur in 'm' ways and another mutually exclusive event 'B' can occur in 'n' ways, then the total number of ways either event 'A' OR event 'B' can occur is m + n.
  • Example: Travelling from City X to City Y by either 2 train routes OR 3 bus routes. Total ways = 2 + 3 = 5.

2. Factorial Notation

• The notation 'n!' (read as 'n factorial') represents the product of the first 'n' natural numbers.
  • n! = n × (n-1) × (n-2) × ... × 3 × 2 × 1
• Key Points:
  • Defined only for non-negative integers.
  • 0! = 1 (by definition, crucial for formulas).
  • 1! = 1
  • n! = n × (n-1)!

3. Permutations (Arrangements)

• A permutation is an arrangement of a number of objects taken some or all at a time, in a definite order. Order matters here!

• Notation: P(n, r) or nPr represents the number of permutations (arrangements) of 'n' distinct objects taken 'r' at a time (where 0 ≤ r ≤ n).

• Formula 1: Permutations of 'n' distinct objects taken 'r' at a time (Repetition NOT allowed):

  • P(n, r) = n! / (n-r)!
  • Think: Filling 'r' places using 'n' distinct items without repetition. The first place has 'n' choices, the second has 'n-1', ..., the r-th has 'n-r+1'. Product = n(n-1)...(n-r+1) = n! / (n-r)!

• Formula 2: Permutations of 'n' distinct objects taken all at a time (Repetition NOT allowed):

  • This is just P(n, n).
  • P(n, n) = n! / (n-n)! = n! / 0! = n!
  • Example: Arranging 5 distinct books on a shelf = 5! ways.

• Formula 3: Permutations when objects are NOT all distinct:

  • The number of permutations of 'n' objects where there are p₁ objects of one kind, p₂ objects of a second kind, ..., pk objects of a k-th kind, such that p₁ + p₂ + ... + pk = n is:
  • n! / (p₁! × p₂! × ... × pk!)
  • Example: Number of distinct arrangements of the letters in the word "ALLAHABAD". (n=9, A=4, L=2, H=1, B=1, D=1). Ways = 9! / (4! × 2! × 1! × 1! × 1!) = 9! / (4! × 2!).

• Formula 4: Permutations when Repetition IS allowed:

  • The number of permutations of 'n' distinct objects taken 'r' at a time when repetition is allowed is:
  • n^r
  • Think: Filling 'r' places using 'n' distinct items with repetition allowed. Each place has 'n' choices. Total ways = n × n × ... × n (r times) = n^r.
  • Example: Forming a 3-digit number using digits 1, 2, 3, 4, 5 with repetition allowed = 5³ = 125.

4. Combinations (Selections)

• A combination is a selection of a number of objects taken some or all at a time, where the order of selection does not matter.

• Notation: C(n, r) or nCr or (n r) represents the number of combinations (selections) of 'n' distinct objects taken 'r' at a time (where 0 ≤ r ≤ n).

• Formula 1: Combinations of 'n' distinct objects taken 'r' at a time:

  • C(n, r) = n! / [r! × (n-r)!]
  • Think: First arrange 'r' items out of 'n' (which is P(n, r)). Since order doesn't matter in combination, divide by the number of ways these 'r' items can be arranged among themselves (which is r!).
  • Hence, C(n, r) = P(n, r) / r!

• Relationship between P(n, r) and C(n, r):

  • P(n, r) = C(n, r) × r!

• Important Properties of C(n, r):

  • C(n, 0) = 1 (Selecting zero objects from n)
  • C(n, n) = 1 (Selecting all n objects from n)
  • C(n, r) = C(n, n-r) (Selecting 'r' objects is the same as rejecting 'n-r' objects). This is very useful for simplification. Example: C(10, 8) = C(10, 10-8) = C(10, 2).
  • Pascal's Rule: C(n, r) + C(n, r-1) = C(n+1, r) (Useful in some proofs and advanced problems).
  • If C(n, x) = C(n, y), then either x = y or x + y = n.

5. Problem-Solving Strategy

1. Read Carefully: Understand what needs to be counted – arrangements or selections? Are objects distinct? Is repetition allowed?
2. Identify Keywords:
   • Permutation keywords: Arrangement, order, schedule, rank, stand in line, digits in a number, letters in a word (usually).
   • Combination keywords: Selection, choose, pick, committee, group, team (usually).
3. Apply FPC: Break down the problem into smaller steps or cases. Use the Multiplication Principle for sequential steps ('AND') and the Addition Principle for mutually exclusive choices ('OR').
4. Use Formulas: Apply the appropriate P(n, r) or C(n, r) formula based on whether order matters and repetition is allowed.
5. Handle Constraints:
   • Items always together: Treat the group of items as a single unit. Arrange this unit with others, then arrange items within the unit.
   • Items never together: Calculate total arrangements - arrangements where they are together.
   • Specific positions: Fix the items in their specified positions first, then arrange the remaining items in the remaining places.
   • At least / At most: Use combinations and consider different cases (e.g., 'at least 2' means 'exactly 2' + 'exactly 3' + ...). Sometimes it's easier to calculate Total - (Unwanted cases).

Example Application (Government Exam Context):

• Problem: In how many ways can a committee of 3 men and 2 women be formed from 7 men and 5 women?
• Solution:
  • Selecting 3 men from 7 (order doesn't matter): C(7, 3) ways.
  • Selecting 2 women from 5 (order doesn't matter): C(5, 2) ways.
  • Since both selections must happen (AND), use the Multiplication Principle.
  • Total ways = C(7, 3) × C(5, 2)
  • C(7, 3) = 7! / (3! × 4!) = (7 × 6 × 5) / (3 × 2 × 1) = 35
  • C(5, 2) = 5! / (2! × 3!) = (5 × 4) / (2 × 1) = 10
  • Total ways = 35 × 10 = 350.

Master these principles and formulas, practice identifying whether a problem requires permutation or combination, and you'll be well-prepared for questions from this topic in your exams.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts we've discussed:

1. The value of P(8, 3) is:
   a) 336
   b) 56
   c) 40320
   d) 6720

2. The value of C(10, 7) is the same as:
   a) C(10, 6)
   b) C(10, 4)
   c) C(10, 3)
   d) P(10, 7)

3. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?
   a) 120
   b) 60
   c) 24
   d) 625

4. In how many ways can a committee of 4 people be selected from a group of 9 people?
   a) 3024
   b) 126
   c) 9! / 4!
   d) 9^4

5. How many distinct permutations of the letters in the word "MISSISSIPPI" are there?
   a) 11!
   b) 11! / (4! * 4! * 2!)
   c) 11! / (4! * 2! * 1!)
   d) C(11, 4) * C(7, 4) * C(3, 2)

6. If C(n, 8) = C(n, 6), then the value of C(n, 2) is:
   a) 14
   b) 28
   c) 91
   d) 56

7. In how many ways can 5 different books be arranged on a shelf?
   a) 25
   b) 5
   c) 120
   d) 60

8. How many 3-letter code words can be formed using the first 10 letters of the English alphabet if repetition of letters is allowed?
   a) P(10, 3)
   b) C(10, 3)
   c) 1000
   d) 720

9. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
   a) C(7,3) + C(7,4) + C(7,5)
   b) C(7,3)C(6,2) + C(7,4)C(6,1) + C(7,5)C(6,0)
   c) C(13, 5) - C(7,0)C(6,5) - C(7,1)C(6,4) - C(7,2)C(6,3)
   d) P(7,3)P(6,2) + P(7,4)P(6,1) + P(7,5)

10. There are 10 points in a plane, no three of which are collinear. How many triangles can be formed by joining these points?
    a) P(10, 3)
    b) 10! / 3!
    c) 720
    d) 120

Answer Key for MCQs:

1. a) 336 (P(8,3) = 8!/(8-3)! = 8!/5! = 876 = 336)
2. c) C(10, 3) (Using C(n, r) = C(n, n-r), C(10, 7) = C(10, 10-7) = C(10, 3))
3. a) 120 (Arrangement without repetition: P(5, 4) = 5!/(5-4)! = 5!/1! = 120)
4. b) 126 (Selection, order doesn't matter: C(9, 4) = 9!/(4!5!) = (9876)/(4321) = 126)
5. b) 11! / (4! * 4! * 2!) (Permutation with non-distinct items: n=11, I=4, S=4, P=2, M=1. Ways = 11!/(4!4!2!1!))
6. c) 91 (If C(n, 8) = C(n, 6), then n = 8+6 = 14. We need C(14, 2) = 14!/(2!12!) = (14*13)/2 = 91)
7. c) 120 (Arrangement of 5 distinct items: P(5, 5) = 5! = 120)
8. c) 1000 (Arrangement with repetition allowed: n=10, r=3. Ways = n^r = 10^3 = 1000)
9. b) C(7,3)C(6,2) + C(7,4)C(6,1) + C(7,5)C(6,0) (Case 1: 3 Men AND 2 Women OR Case 2: 4 Men AND 1 Woman OR Case 3: 5 Men AND 0 Women. Use Combination and FPC.)
10. d) 120 (Selecting 3 non-collinear points out of 10 forms a triangle. Order doesn't matter: C(10, 3) = 10!/(3!7!) = (1098)/(321) = 120)

Make sure you understand the reasoning behind each answer. Keep practicing!