Class 11 Mathematics Notes Chapter 7 (Permutations and combinations) – Mathematics Book

Detailed Notes with MCQs of Chapter 7: Permutations and Combinations. This is a fundamental topic in mathematics, especially important for competitive government exams as it tests your logical reasoning and counting skills. Mastering the concepts here will significantly help in probability as well.

We are essentially learning how to count the number of ways certain events can occur, either through arrangements or selections.

Chapter 7: Permutations and Combinations - Detailed Notes

1. Fundamental Principle of Counting (FPC)

This is the bedrock of the entire chapter. It has two main parts:

• Multiplication Principle: If an event 'A' can occur in 'm' different ways, and following that, another independent event 'B' can occur in 'n' different ways, then the total number of ways that both events A and B can occur in the stated order is m × n.
  • Example: If you have 3 shirts and 2 pairs of trousers, you can make 3 × 2 = 6 different outfits.
  • This principle can be extended to any finite number of events.
• 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 that either event A or event B can occur is m + n.
  • Example: If you want to travel from City X to City Y, and there are 3 bus routes and 2 train routes (and you can only take one), you have 3 + 2 = 5 choices for your travel.

2. Factorial Notation

For a natural number 'n', the factorial of n, denoted by n! or |n, is the product of the first n natural numbers.
n! = n × (n - 1) × (n - 2) × ... × 3 × 2 × 1

• Example: 5! = 5 × 4 × 3 × 2 × 1 = 120
• Key Property: 0! = 1 (This is defined for convenience and consistency in formulas).
• Recursive Property: n! = n × (n - 1)!

3. Permutations

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

• a) Permutations of 'n' distinct objects taken 'r' at a time (n ≥ r):
  • This is denoted by P(n, r) or ⁿP<0xE1><0xB5><0xA3>.
  • Formula: P(n, r) = n! / (n - r)!
  • Logic: You have 'n' choices for the 1st position, 'n-1' for the 2nd, ..., 'n-r+1' for the r-th position. Multiplying these gives n × (n-1) × ... × (n-r+1), which simplifies to the formula above.
  • Example: How many 3-letter words (with or without meaning) can be formed from the letters of the word 'NUMBER' without repetition? Here n=6, r=3. P(6, 3) = 6! / (6-3)! = 6! / 3! = (6 × 5 × 4 × 3!) / 3! = 6 × 5 × 4 = 120.
• b) Permutations of 'n' distinct objects taken all at a time:
  • This is P(n, n) = n! / (n - n)! = n! / 0! = n! / 1 = n!
  • Example: How many ways can 5 people stand in a queue? 5! = 120 ways.
• c) Permutations when repetition is allowed:
  • The number of permutations of 'n' distinct objects taken 'r' at a time, where repetition is allowed, is nʳ.
  • Logic: You have 'n' choices for the 1st position, 'n' choices for the 2nd (since repetition is allowed), ..., 'n' choices for the r-th position. Total ways = n × n × ... × n (r times) = nʳ.
  • Example: How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 if repetition is allowed? Here n=5, r=3. Number of ways = 5³ = 125.
• d) 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, ..., p<0xE2><0x82><0x96> objects of a k-th kind, such that p₁ + p₂ + ... + p<0xE2><0x82><0x96> = n, is given by:
    n! / (p₁! × p₂! × ... × p<0xE2><0x82><0x96>!)
  • Example: How many distinct arrangements of the letters in the word 'MISSISSIPPI'? Here n=11. M=1, I=4, S=4, P=2. Number of arrangements = 11! / (1! × 4! × 4! × 2!) = 34650.

4. Combinations

A combination is a selection of a number of objects taken some or all at a time. Order does not matter here!

• a) Combinations of 'n' distinct objects taken 'r' at a time (n ≥ r):
  • This is denoted by C(n, r) or ⁿC<0xE1><0xB5><0xA3> or (ⁿ<0xE1><0xB5><0xA3>).
  • Formula: C(n, r) = n! / (r! × (n - r)!)
  • Logic: The number of permutations P(n, r) involves selecting 'r' objects and arranging them. The arrangement part contributes r! ways. So, P(n, r) = C(n, r) × r!. Rearranging this gives the formula for C(n, r).
  • Example: How many ways can a committee of 3 people be selected from a group of 5 people? Here n=5, r=3. Order doesn't matter in a committee. C(5, 3) = 5! / (3! × (5-3)!) = 5! / (3! × 2!) = (5 × 4 × 3!) / (3! × 2 × 1) = (5 × 4) / 2 = 10.
• b) Important Properties of Combinations:
  • C(n, r) = C(n, n - r) (Example: C(10, 8) = C(10, 10-8) = C(10, 2))
  • C(n, 0) = 1 (Selecting nothing is one way)
  • C(n, n) = 1 (Selecting all items is one way)
  • C(n, 1) = n
  • Pascal's Rule: C(n, r) + C(n, r-1) = C(n+1, r)

5. Key Difference: Permutation vs. Combination

• Use Permutation when the order of items is important (e.g., arranging letters in a word, assigning specific roles like President/VP, forming numbers). Keywords: Arrangement, Order, Schedule, Rank.
• Use Combination when the order of items is irrelevant (e.g., selecting a team or committee, choosing items from a menu, picking cards from a deck). Keywords: Selection, Group, Committee, Team, Choose, Pick.

Summary Table

Remember to carefully read the problem statement in exams to determine whether order matters or not before applying the correct formula. Practice is key!

Multiple Choice Questions (MCQs) for Practice

1. In how many ways can 3 prizes be distributed among 5 students, if no student gets more than one prize?
   (A) 10
   (B) 60
   (C) 125
   (D) 15
   Answer: (B) (Explanation: Order matters as prizes are distinct. P(5, 3) = 5!/(5-3)! = 5!/2! = 120/2 = 60)

2. The value of C(10, 8) is:
   (A) 10
   (B) 90
   (C) 45
   (D) 180
   Answer: (C) (Explanation: C(10, 8) = C(10, 10-8) = C(10, 2) = 10! / (2! * 8!) = (10 × 9) / (2 × 1) = 45)

3. How many different words can be formed using all the letters of the word 'EQUATION'?
   (A) 8! / 2!
   (B) 8!
   (C) C(8, 8)
   (D) 8⁸
   Answer: (B) (Explanation: All 8 letters are distinct. Number of arrangements = P(8, 8) = 8!)

4. How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?
   (A) P(5, 4)
   (B) C(5, 4)
   (C) 5⁴
   (D) 4⁵
   Answer: (A) (Explanation: Order matters in forming a number, and repetition is not allowed. n=5, r=4. Number of ways = P(5, 4) = 5!/(5-4)! = 5! = 120)

5. In how many ways can a committee of 3 men and 2 women be formed from 7 men and 5 women?
   (A) P(7,3) × P(5,2)
   (B) C(7,3) + C(5,2)
   (C) C(12, 5)
   (D) C(7,3) × C(5,2)
   Answer: (D) (Explanation: Select 3 men from 7 (order doesn't matter) AND select 2 women from 5 (order doesn't matter). Use FPC (Multiplication). Ways = C(7,3) × C(5,2) = [7!/(3!4!)] × [5!/(2!3!)] = 35 × 10 = 350)

6. How many distinct arrangements of the letters in the word 'LEADER' can be made?
   (A) 720
   (B) 360
   (C) 120
   (D) 6! / 2! 2!
   Answer: (B) (Explanation: n=6. L=1, E=2, A=1, D=1, R=1. Number of arrangements = 6! / 2! = 720 / 2 = 360)

7. If C(n, 9) = C(n, 8), find the value of n.
   (A) 9
   (B) 8
   (C) 17
   (D) 1
   Answer: (C) (Explanation: Using the property C(n, r) = C(n, n-r). If C(n, 9) = C(n, 8), then either 9 = 8 (not possible) or 9 = n - 8. Therefore, n = 9 + 8 = 17. Alternatively, use C(n, r) + C(n, r-1) = C(n+1, r). Here, C(n,9)=C(n,8) implies using C(n,r) = C(n, n-r), so n-9=8 or n=17)

8. How many chords can be drawn through 21 points on a circle?
   (A) P(21, 2)
   (B) 21!
   (C) C(21, 2)
   (D) 210
   Answer: (C) (Explanation: A chord is formed by selecting any 2 points from the 21 points. Order doesn't matter. Number of chords = C(21, 2) = 21! / (2! * 19!) = (21 × 20) / 2 = 210. Note: Option D gives the value, but C asks for the expression)

9. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
   (A) 120
   (B) 108
   (C) 216
   (D) 60
   Answer: (B) (Explanation: For the number to be even, the unit's place must be 2, 4, or 6 (3 choices). Since repetition is allowed, the hundred's place can be filled in 6 ways and the ten's place can be filled in 6 ways. Total numbers = 6 × 6 × 3 = 108)

10. What is the value of 0! + 1! + 2!?
    (A) 3
    (B) 4
    (C) 2
    (D) 0
    Answer: (B) (Explanation: 0! = 1, 1! = 1, 2! = 2. Sum = 1 + 1 + 2 = 4)

Keep practicing these concepts with various problems. Good luck with your preparation!