Board exam focus — Permutations and Combinations (CBSE & HBSE)

Combinatorial reasoning chapter — sets up probability and binomial theorem. CBSE expects 6–8 marks: factorial computation, permutations with restrictions, combinations, and a word problem on arrangements/selections.

Fundamental Counting Principle & Factorial

Fundamental Principle of Counting (Multiplication)

If an event can occur in m ways and a second independent event in n ways, the two events can occur in succession in m × n ways.

Generalisation: for k independent stages with n₁, n₂, …, nₖ outcomes, total outcomes = n₁ × n₂ × … × nₖ.

Addition Principle

If two events are mutually exclusive (cannot happen together) with m and n ways respectively, total = m + n.

Factorial Notation

n! = n × (n − 1) × (n − 2) × … × 2 × 1.

Conventions:

• 0! = 1
• 1! = 1
• n! = n × (n − 1)!
• (n − r)! divides n! evenly for r ≤ n.

Examples of Quick Factorial

When to Use Which

Useful Factorial Manipulations

• (n + 1)! = (n + 1) × n!
• n!/(n − r)! = n × (n − 1) × … × (n − r + 1) (product of r terms)
• n!/r! · (n − r)! = ⁿCᵣ (binomial coefficient — next module)

CBSE Tip: When evaluating large factorials, cancel before multiplying — e.g. 10!/8! = 10 × 9 = 90, not 3628800 / 40320.

Permutations — Order Matters

Definition

Permutation of n different objects taken r at a time = number of arrangements where order matters.

Notation: ⁿPᵣ or P(n, r) or P_r^n.

Formula

ⁿPᵣ = n! / (n − r)! for 0 ≤ r ≤ n.

Special cases:

• ⁿP₀ = 1 (only one arrangement of nothing — the empty arrangement)
• ⁿP₁ = n
• ⁿPₙ = n! (arrange all n)

Derivation

For the first position: n choices. For the second: n − 1 (already used one). … For the r-th: n − r + 1. Product = n × (n − 1) × … × (n − r + 1) = n!/(n − r)!.

Permutations with Repetition

When objects can repeat: Number of r-letter words from n distinct letters = nʳ.

Permutations of n Objects Not All Distinct

If there are n objects with p alike of one kind, q alike of another, etc., the number of distinct arrangements is: n! / (p! × q! × …).

Circular Permutations

Number of arrangements of n distinct objects around a circle = (n − 1)!. (Divide by 2 if clockwise and anti-clockwise are considered the same — like a necklace.)

Common Restrictions

• Certain objects always together: treat them as one block, arrange the block, then arrange within.
• Certain objects never together: total − (always together).
• Specific positions fixed: arrange remaining around fixed.

HBSE Tip: Read the question carefully — "different orderings" indicates permutations; "selections" indicates combinations.

Combinations — Order Doesn't Matter

Definition

Combination of n distinct objects taken r at a time = number of selections (order irrelevant).

Notation: ⁿCᵣ or C(n, r) or $\binom{n}{r}$.

Formula

ⁿCᵣ = n! / (r! · (n − r)!) for 0 ≤ r ≤ n.

Link with permutations: ⁿPᵣ = ⁿCᵣ × r!.

Useful Identities

1. ⁿC₀ = ⁿCₙ = 1
2. ⁿC₁ = n
3. ⁿCᵣ = ⁿCₙ₋ᵣ (symmetry)
4. ⁿCᵣ + ⁿCᵣ₋₁ = (n+1)Cᵣ (Pascal's rule)
5. ⁿCᵣ / ⁿCᵣ₋₁ = (n − r + 1)/r
6. Σⁿᵣ₌₀ ⁿCᵣ = 2ⁿ

Geometric Interpretations

• Number of diagonals of an n-gon = ⁿC₂ − n.
• Number of triangles from n points (no three collinear) = ⁿC₃.
• Number of lines from n points = ⁿC₂.

When to Use Combination vs Permutation

Mixed Problems

Many problems first select (combination), then arrange (permutation) the selected items. The total is the product.

Example: From 10 players, form a team of 5 and arrange them in batting order:

• Select 5: ¹⁰C₅ = 252.
• Arrange: 5! = 120.
• Total: 252 × 120 = 30240.

CBSE Tip: ⁿCᵣ = ⁿCₙ₋ᵣ saves time — choose the smaller subscript (e.g. ¹⁰C₇ = ¹⁰C₃).

Mixed Problems & Identities

Strategy for Compound Problems

1. Read the question for cues: arrange / select / together / not together / at least / at most.
2. Break the problem into stages.
3. Apply multiplication for stages performed in sequence; addition for mutually exclusive cases.
4. Use complement when "at least one" is involved (total − "none").

"At Least" and "At Most"

At least k ⇒ sum from k to max. At most k ⇒ sum from 0 to k. Use complement when easier: P(at least 1) = 1 − P(none).

Distribution Problems

• Distinct balls in distinct boxes: nᵏ (k balls, n boxes).
• Identical balls in distinct boxes: stars-and-bars (not in NCERT 11, optional).

Sum of Binomial Coefficients

• ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ.
• ⁿC₀ + ⁿC₂ + … = ⁿC₁ + ⁿC₃ + … = 2ⁿ⁻¹.

Pascal's Triangle Construction

Row n consists of ⁿC₀, ⁿC₁, …, ⁿCₙ. Each entry = sum of two entries above it.

Common Word-Problem Patterns

Classic Examples

• Number of handshakes among n people = ⁿC₂.
• Number of diagonals of an n-gon = ⁿC₂ − n.
• Number of lines through n points (no 3 collinear) = ⁿC₂.

HBSE Tip: For arrangements with restrictions, draw a tiny diagram showing fixed positions and gaps before computing.

Frequently asked questions

Are these Permutations and Combinations notes free?

Yes — the Permutations and Combinations notes for Mathematics (Class 11) on Siksha Sarovar are completely free to read, with no account required.

Do these notes follow CBSE and HBSE?

Yes. The Permutations and Combinations notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.

What does the Permutations and Combinations chapter cover?

Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Permutations and Combinations.