Probability &
Combination

Core formulas · key identities · exam-ready worked examples

1Fundamental Counting Principle

Events A(m ways) × B(n ways) Total = m × n × ···

Multiply independent choices together.

2Permutation ⁿPᵣ

ⁿPᵣ = n! / (n−r)! = n(n−1)···(n−r+1) ⁿPₙ = n!

Order MATTERS — selecting r from n distinct objects.

3Combination ⁿCᵣ

ⁿCᵣ = n! / (r! × (n−r)!) ⁿCᵣ = ⁿCₙ₋ᵣ (symmetry) ⁿC₀ = ⁿCₙ = 1

Order does NOT matter — choosing r from n objects.

4Repeated-Letter Arrangements

n letters with a alike, b alike, c alike, …: n! / (a! × b! × c! × …)

Divide by the factorial of each repeated group.

5Combination with Repetition

Choose r items from n types (repetition allowed): H(n,r) = ⁿ⁺ʳ⁻¹Cᵣ

Also called multiset coefficient.

6Key Identities

Pascal: ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁ Sum: 2ⁿ = Σ ⁿCᵣ (r=0 to n) Circle: (n−1)! for n objects

Essential identities for advanced problems.

Must-Memorize Facts

PPermutation (order matters): ⁿPᵣ = n!/(n−r)! | 6P3 = 6×5×4 = 120

CCombination (order doesn't matter): ⁿCᵣ = n!/r!(n−r)! | 7C3 = 35

SYMSymmetry: ⁿCᵣ = ⁿCₙ₋ᵣ | 12C9 = 12C3 = 220

SUMSum rule: ⁿC₀+ⁿC₁+···+ⁿCₙ = 2ⁿ | n=5 → 32

REPRepeated items: SERIES → 6!/(2!·2!) = 180

COMPComplement: "at least 1 X" = Total − "no X" (complement method)

LINELines from points: no 3 collinear → ⁿC₂ lines

CIRCCircular permutation: (n−1)! for n distinct objects around a circle

Worked Examples

Example 1 — Permutation

How many ways can President, Vice-President, and Secretary be chosen from 10 students?

¹⁰P₃ = 10 × 9 × 8 = 720 ways

Example 2 — Combination

How many ways can a committee of 4 be chosen from 9 people?

⁹C₄ = (9×8×7×6)/(4×3×2×1) = 3024/24 = 126 ways

Example 3 — Complement Method

From 4 men and 3 women, how many 3-person committees include at least 1 woman?

Total = ⁷C₃ = 35 ; All-male = ⁴C₃ = 4 Answer = 35 − 4 = 31 ways

Example 4 — Repeated Letters

How many distinct arrangements are there of the letters in "LEVEL"?

L(2), E(2), V(1) — 5 letters total Answer = 5!/(2!×2!) = 120/4 = 30 arrangements

Example 5 — Points and Lines

8 points in a plane, no 3 collinear. How many straight lines?

Each line needs exactly 2 points. Answer = ⁸C₂ = 28 lines

Probability &Combination