Permutations and Combinations Class 11 Maths Chapter Notes

Key Terms and Definitions

Factorial Notation (n!)

The product of the first n positive integers. It is denoted by n!. So, n! = 1 × 2 × 3 × ... × n. By definition, 0! = 1.

Permutation (P)

An arrangement of a number of objects in a definite order. The number of permutations of 'n' distinct objects taken 'r' at a time is denoted by nPr.

Combination (C)

A selection of a number of objects where the order of selection does not matter. The number of combinations of 'n' distinct objects taken 'r' at a time is denoted by nCr.

Fundamental Principle of Multiplication

If an event can occur in 'm' different ways, and another independent event can occur in 'n' different ways, then the total number of ways both events can occur in succession is m × n.

Fundamental Principle of Addition

If an event can be done in 'm' ways and another mutually exclusive event can be done in 'n' ways, then either of the two can be done in (m + n) ways.

Circular Permutation

The arrangement of objects in a circle. The number of ways to arrange 'n' distinct objects in a circle is (n-1)!.

The Foundation: Fundamental Principle of Counting

Everything in this chapter builds upon the Fundamental Principle of Counting. It's split into two key ideas: multiplication and addition.

1. Multiplication Principle (The 'AND' Rule): This is used when you need to perform a series of tasks one after the other. Think of it as a sequence of choices. If you have Task A that can be done in m ways, AND then Task B can be done in n ways, the total number of ways to complete the sequence (Task A followed by Task B) is m × n. For example, if you have 3 shirts and 2 pairs of trousers, the total number of different outfits you can create is 3 × 2 = 6. You choose a shirt and you choose a trouser.

1. Addition Principle (The 'OR' Rule): This is used when you have a choice between mutually exclusive options. If you can perform Task A in m ways OR Task B in n ways, and you cannot do both at the same time, then the total number of ways to perform either task is m + n. For example, if a library has 5 books on Physics and 4 books on Chemistry, and you need to choose just one book, the number of ways you can choose a book is 5 + 4 = 9. You choose a Physics book or a Chemistry book.

Must-Remember Formulas and Properties

• Factorial: n! = n × (n-1) × ... × 2 × 1. Remember: 0! = 1.
• Permutation Formula (nPr): The number of arrangements of 'n' objects taken 'r' at a time is nPr = n! / (n-r)! where 0 ≤ r ≤ n.
• Combination Formula (nCr): The number of selections of 'n' objects taken 'r' at a time is nCr = n! / (r! * (n-r)!) where 0 ≤ r ≤ n.
• Relationship between nPr and nCr: nPr = nCr × r!
• Important Combination Properties: nCr = nCn-r, nC0 = 1, nCn = 1.
• Pascal's Rule: nCr + nC(r-1) = (n+1)Cr. This is useful for simplifying expressions.
• Permutations with Repetition: The number of permutations of 'n' objects where p1 objects are of one kind, p2 are of a second kind, ..., pk are of a k-th kind is n! / (p1! p2! ... * pk!). (Think: MISSISSIPPI).
• Circular Permutations: The number of ways to arrange 'n' distinct objects around a circle is (n-1)!.
• Sum of Binomial Coefficients: The sum of all combinations from a set of n items is nC0 + nC1 + nC2 + ... + nCn = 2^n.

Permutation vs. Combination: The Core Difference

Worked Examples

• {"title":"Example 1: Basic Permutation","bodyMarkdown":"Problem: In how many ways can 5 people be seated in a row on 3 empty chairs?\n\nSolution:\nHere, we are arranging 5 people into 3 positions. Order matters.\nThis is a permutation of 5 objects taken 3 at a time.\nn = 5, r = 3\n5P3 = 5! / (5-3)! = 5! / 2! = (5 × 4 × 3 × 2!) / 2! = 5 × 4 × 3 = 60\nAnswer: 60 ways."}
• {"title":"Example 2: Basic Combination","bodyMarkdown":"Problem: A committee of 3 members is to be formed from a group of 8 people. In how many ways can it be done?\n\nSolution:\nHere, we are selecting a group of 3 from 8. The order in which members are chosen does not matter.\nThis is a combination of 8 objects taken 3 at a time.\nn = 8, r = 3\n8C3 = 8! / (3! (8-3)!) = 8! / (3! 5!) = (8 × 7 × 6 × 5!) / ((3 × 2 × 1) * 5!) = 8 × 7 = 56\nAnswer: 56 ways."}
• {"title":"Example 3: Permutation with Repetition","bodyMarkdown":"Problem: Find the number of distinct arrangements of the letters in the word 'SUCCESS'.\n\nSolution:\nThe word 'SUCCESS' has 7 letters in total (n=7).\nThe letter 'S' is repeated 3 times (p1=3).\nThe letter 'C' is repeated 2 times (p2=2).\nThe letters 'U' and 'E' occur once.\nUsing the formula for permutations with repetition:\nNumber of arrangements = n! / (p1! p2!) = 7! / (3! 2!) = (7 × 6 × 5 × 4 × 3!) / (3! × 2 × 1) = 7 × 6 × 5 × 2 = 420\nAnswer: 420 distinct arrangements."}

Exam Traps & Scoring Tips

1. P vs. C Confusion: This is the most common trap. Before applying a formula, ask yourself: "Does the order matter here?". If the question involves arranging, ranking, or assigning to specific roles (President, VP), use Permutation (P). If it involves selecting, choosing, or forming a group/committee, use Combination (C).

2. 'At Least' / 'At Most' Problems: These questions often require breaking the problem into cases and using the Addition Principle. For example, 'at least 2' means '(exactly 2) OR (exactly 3) OR ...'. Calculate each case and add them up.

3. 'Together' vs. 'Never Together':

• Items Together: Treat the items that must be together as a single block or unit. Permute this block with other items, and then multiply by the internal permutations of the items within the block.
• Items Never Together: Use the subtraction principle. Calculate the Total arrangements (without any restrictions) and subtract the Arrangements where the items are always together. This is often easier than calculating 'never together' directly.

4. Don't Forget 0! = 1: In calculations like 5P5 = 5! / (5-5)! = 5! / 0!, remember that 0! is 1, not 0. This is a common calculation error.

Quick Revision Check

• Q: Calculate 10C8. A: Using the property nCr = nCn-r, 10C8 = 10C(10-8) = 10C2 = (10 9) / (2 1) = 45.
• Q: How many 4-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if repetition of digits is not allowed? A: This is an arrangement of 5 digits taken 4 at a time. 5P4 = 5! / (5-4)! = 5! / 1! = 120.
• Q: A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has no girls? A: We need to select 5 boys from 7 boys. This is a selection problem. 7C5 = 7C2 = (7 6) / (2 1) = 21.
• Q: In how many ways can 6 people be arranged around a circular table? A: Using the circular permutation formula (n-1)!, we get (6-1)! = 5! = 120.

Frequently Asked Questions

Frequently Asked Questions