Permutations and Combinations – Class 11 Maths Chapter 6 Notes & NCERT PDF

Class 11MathematicsChapter 6NCERT book: Mathematics - Class 11

Notes and a simple summary of Chapter 6 of the NCERT Class 11 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 6: Permutations and Combinations

Introduction

A suitcase has a 4-wheel number lock (digits 0–9, no repetition) and you remember only that the first digit is 7. How many 3-digit sequences must you try? Listing them is tedious — the answer is the number of arrangements of the 9 remaining digits taken 3 at a time, ⁹P₃ = 9 × 8 × 7 = 504. This chapter builds the counting tools — the fundamental principle of counting, permutations (arrangements, order matters) and combinations (selections, order does not matter) — that let us count without listing.

Key Concepts & Theorems

6.2 Fundamental Principle of Counting (multiplication principle)

If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrences of the events in the given order is m × n.

It extends to any number of events (m × n × p …). Mohan with 3 pants and 2 shirts has 3 × 2 = 6 outfits; Sabnam with 2 bags, 3 tiffin boxes and 2 bottles has 2 × 3 × 2 = 12 ways.

6.3 Permutations

Definition: A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.

Factorial notation: n! = 1 × 2 × 3 × … × n, with 0! = 1 and n! = n × (n − 1)!.

Theorem 1: Number of permutations of n different objects taken r at a time (0 < r ≤ n), repetition not allowed: ⁿPᵣ = n(n − 1)(n − 2)…(n − r + 1) = n!/(n − r)!. Also ⁿP₀ = 1 and ⁿPₙ = n!.

Theorem 2: With repetition allowed: nʳ.

Theorem 3: Permutations of n objects where p are alike and the rest different: n!/p!.

Theorem 4: Where p₁ are of one kind, p₂ of a second kind, …, pₖ of a k-th kind: n!/(p₁! p₂! … pₖ!).

6.4 Combinations

Definition: A selection where the order is not important is a combination. The number of combinations of n different objects taken r at a time is ⁿCᵣ.

Theorem 5: ⁿPᵣ = ⁿCᵣ × r!, so ⁿCᵣ = n!/[r!(n − r)!], 0 ≤ r ≤ n.

Remarks: ⁿC₀ = 1, ⁿCₙ = 1, and ⁿCᵣ = ⁿCₙ₋ᵣ (selecting r objects is the same as rejecting n − r). Hence if ⁿCₐ = ⁿC_b then a = b or a + b = n.

Theorem 6: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.

Worked Examples (from the chapter)

Example 1: 4-letter words from ROSE, no repetition: 4 × 3 × 2 × 1 = 24; with repetition 4⁴ = 256.

Example 4: Signals of at least 2 flags from 5 different flags: 20 + 60 + 120 + 120 = 320.

Example 9: Arrangements of ALLAHABAD (9 letters: 4 A's, 2 L's): 9!/(4! 2!) = 7560.

Example 11: 3-digit numbers (between 100 and 1000) from 0, 1, 2, 3, 4, 5 without repetition: ⁶P₃ − ⁵P₂ = 120 − 20 = 100 (remove those with 0 in the hundreds place).

Example 12: ⁿP₅ = 42 ⁿP₃ ⇒ (n − 3)(n − 4) = 42 ⇒ n = 10.

Example 14: DAUGHTER — vowels A, U, E together: treat them as one block: 6! × 3! = 4320; vowels not all together: 8! − 4320 = 36000.

Example 16: INDEPENDENCE (12 letters: N 3, E 4, D 2): 12!/(3! 4! 2!) = 1663200.

Example 17: ⁿC₉ = ⁿC₈ ⇒ n = 17 ⇒ ¹⁷C₁₇ = 1.

Example 19: 4 cards from 52: ⁵²C₄ = 270725; all from the same suit: 4 × ¹³C₄ = 2860; one from each suit: 13⁴ = 28561.

Example 21: Team of 5 from 4 girls and 7 boys: no girl ⁷C₅ = 21; at least one boy and one girl = 7 + 84 + 210 + 140 = 441; at least 3 girls = 84 + 7 = 91.

Example 24: 5 girls and 3 boys in a row, no two boys together: seat girls in 5! ways, then boys in the 6 gaps in ⁶P₃ ways: 120 × 120 = 14400.

Formula Sheet

Multiplication principle: m × n × p …

n! = n(n − 1)!, 0! = 1

ⁿPᵣ = n!/(n − r)!; with repetition nʳ; ⁿPₙ = n!

Alike objects: n!/(p₁! p₂! … pₖ!)

ⁿCᵣ = n!/[r!(n − r)!] = ⁿPᵣ/r!

ⁿCᵣ = ⁿCₙ₋ᵣ; ⁿC₀ = ⁿCₙ = 1

ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

Exam Focus

Decide order matters? → P; order does not matter? → C. Committees, teams, card hands and chords are combinations; words, numbers, signals and seating are permutations.

"Together" → glue the objects into one block, then multiply by their internal arrangements. "Not together" → total − together, or use the gap method (Example 24).

"At least" → add the separate cases (Example 21).

Watch zero in number problems: subtract arrangements with 0 in the leading place.

Answer outline — "How many words from INVOLUTE with 3 vowels and 2 consonants?": ⁴C₃ × ⁴C₂ = 24 selections, each arranged in 5! ways ⇒ 24 × 120 = 2880.

Answer outline — "Find r if 5 ⁴Pᵣ = 6 ⁵Pᵣ₋₁": simplify to (6 − r)(5 − r) = 6 ⇒ r² − 11r + 24 = 0 ⇒ r = 3 or 8; since r ≤ 4, r = 3.

Historical Note (as the book tells it)

The Jains treated the subject as Vikalpa; Mahavira (around 850) gave general formulae for permutations and combinations. Sushruta (6th century B.C.) noted that 63 combinations can be made from 6 tastes; Pingala (around 3rd century B.C.) counted combinations of letters in Chhanda Sutra; Bhaskaracharya (born 1114) treated it as Anka Pasha in Lilavati. The first complete treatment is Ars Conjectandi by Jacob Bernoulli, published in 1713.

Key Terms

Fundamental principle of counting (गणना का आधारभूत सिद्धांत)

Permutation (क्रमचय) — arrangement; order matters

Combination (संचय) — selection; order does not matter

Factorial (क्रमगुणित) — n!

Repetition (पुनरावृत्ति)

Alike objects (समरूप वस्तुएँ)

Arrangement (विन्यास)

Selection (चयन)

Vowel / consonant (स्वर / व्यंजन)

Signal (संकेत) — flags in order

Committee (समिति) — a combination

Common Mistakes

Using ⁿCᵣ when order matters (or ⁿPᵣ when it does not).

Forgetting 0! = 1, or writing (3! + 4!) as 7! (it is 30, not 5040).

Not removing numbers that start with 0.

For "not together", subtracting incorrectly or forgetting the internal arrangement of the glued block.

Double counting in "at least" problems by picking one item first and then "any" for the rest.

Forgetting to divide by the factorials of repeated letters.

💡 Key Learning Points

✓Apply the fundamental principle of counting (m × n × p …) to successive events.
✓Use factorial notation and the formula ⁿPᵣ = n!/(n − r)! for arrangements of distinct objects without repetition, and nʳ with repetition.
✓Count arrangements with alike objects using n!/(p₁! p₂! … pₖ!).
✓Distinguish selection from arrangement and use ⁿCᵣ = n!/[r!(n − r)!], ⁿCᵣ = ⁿCₙ₋ᵣ and ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ.
✓Solve mixed problems involving 'together', 'not together', 'at least' and restrictions on zero.

👨‍🏫 Teaching Tips

→Start with the forgotten-lock story and let students estimate before revealing ⁹P₃ = 504.
→Draw tree diagrams for Mohan's outfits (3 × 2) before stating the multiplication principle.
→Use physical letter cards (e.g. ROSE, AGAIN) so students see why repeated letters reduce the count.
→Hold a 'P or C?' sorting game with situations: committees, passwords, handshakes, seating, lottery tickets.
→Teach the block method and the gap method side by side with seating demonstrations in class.
→Show ⁴P₂ = ⁴C₂ × 2! by listing AB, AC, … then their reverses, to motivate Theorem 5.
→End with the Indian historical note (Vikalpa, Sushruta's 63 tastes, Pingala, Bhaskaracharya).

📋 Assessment Questions

Can the student decide whether a situation is a permutation or a combination and justify it?
Can the student evaluate ⁿPᵣ and ⁿCᵣ and simplify factorial expressions?
Can the student count arrangements of a word with repeated letters?
Can the student handle restrictions (zero not first, items together or apart)?
Can the student split an 'at least' selection problem into cases and add them?

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