Permutations and Combinations Explained

🕐 11 min read | Class 10–12 | FBISE · CBSE · IGCSE · O-Levels · IB

How many ways can five runners finish a race in first, second, and third place? How many three-person committees can be chosen from a group of ten? Both questions involve selecting from a group — but the first problem cares about order (gold, silver, bronze are different outcomes), while the second does not (the same three people form the same committee regardless of which order they were chosen). This single distinction — whether order matters — separates permutations from combinations, and it is the key to unlocking an entire branch of mathematics.

Factorial Notation — The Foundation

Before permutations and combinations, you need to understand factorial notation. The factorial of a positive integer n (written n!) is the product of all positive integers from 1 up to n.

n! = n × (n−1) × (n−2) × … × 2 × 1

Examples: 4! = 4×3×2×1 = 24 | 5! = 120 | 0! = 1 (by definition)

Factorial counts the number of ways to arrange n distinct items in a line. 4 books can be arranged on a shelf in 4! = 24 different orders.

Permutations — When Order Matters

A permutation is an arrangement of items where the order of selection matters. The number of ways to arrange r items chosen from n distinct items is:

ⁿPᵣ = n! / (n − r)!

Step-by-Step Example — Permutation

In a race with 8 runners, how many ways can the gold, silver, and bronze medals be awarded?

📋 Calculating ⁸P₃

1

Identify n and r: n = 8 runners, r = 3 positions. Order matters (1st ≠ 2nd ≠ 3rd).

2

Apply the formula:

⁸P₃ = 8! / (8−3)! = 8! / 5!

3

Simplify:

= (8 × 7 × 6 × 5!) / 5! = 8 × 7 × 6 = 336

There are 336 different ways to award the three medals from 8 runners.

Combinations — When Order Does Not Matter

A combination is a selection of items where the order does not matter. Choosing team members, picking lottery numbers, or forming a committee — none of these depend on order. The formula divides the permutation count by r! to remove the overcounting caused by different orderings of the same group:

ⁿCᵣ = n! / [r! × (n − r)!]

Also written as C(n, r) or "n choose r"

Step-by-Step Example — Combination

A school selects 3 students from a class of 10 to represent the school at a conference. How many ways can this group be chosen?

📋 Calculating ¹⁰C₃

1

Identify n and r: n = 10 students, r = 3. Order does not matter (the same three students form the same group).

2

Apply the formula:

¹⁰C₃ = 10! / (3! × 7!)

3

Simplify:

= (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120

There are 120 different ways to choose the 3 students. Notice this is much less than ⁱ⁰P₃ = 720, because combinations don't count different orderings of the same group.

Permutation vs Combination — The Key Question

💡 Ask yourself: "Would swapping two selected items give a different outcome?" If yes → Permutation. If no → Combination.

Scenario	Order Matters?	Use
Medal positions in a race	Yes	Permutation
Arranging books on a shelf	Yes	Permutation
PIN codes and passwords	Yes	Permutation
Choosing a committee	No	Combination
Selecting lottery numbers	No	Combination
Picking a group to travel	No	Combination

Real-Life Applications

🔐
Cybersecurity: Password strength is calculated using permutations — a 6-character PIN using digits 0–9 has 10⁶ = 1,000,000 possibilities because order matters.
🃏
Card games and probability: The number of possible 5-card poker hands is ⁵²C₅ = 2,598,960. Combinations underpin all card game probability calculations.
💊
Medicine: When testing drug combinations, researchers use combinations to count how many distinct pairings or triplings of n candidate drugs are possible.
🏆
Sports scheduling: The number of unique matches in a round-robin tournament with n teams is ⁿC₂ — combinations, because Team A vs Team B is the same match as Team B vs Team A.

Common Mistakes Students Make

⚠️ Using permutation when combination is needed (or vice versa). Always identify whether order matters before applying any formula. This is the single most important step.

⚠️ Forgetting that 0! = 1. This comes up when r = n (selecting all items). ⁿCₙ = n!/(n! × 0!) = 1, which makes sense — there is exactly one way to choose all items.

⚠️ Not cancelling factorials before multiplying. Always simplify before multiplying out. For ¹⁰C₃, cancel the 7! rather than computing 10! in full.

⚠️ Confusing "at least" and "exactly." Questions asking for "at least 2 from a group" require adding the probabilities of 2, 3, 4 … up to n separately, or using the complement rule.

Frequently Asked Questions

"n choose r" is another way of saying ⁿCᵣ — the number of ways to choose r items from n without regard to order. It is read aloud as "n choose r." The notation C(n, r), ⁿCᵣ, and the binomial coefficient symbol are all equivalent.

Yes, always. The combination formula always produces a whole number because it counts discrete arrangements. If your calculation gives a non-integer, you have made an arithmetic error.

The entries in Pascal's triangle are exactly the combination values ⁿCᵣ. The nth row of Pascal's triangle gives ⁿC₀, ⁿC₁, ⁿC₂, … ⁿCₙ. This connection is fundamental in binomial expansion and is tested in IB and A-Level Mathematics.

Yes. Permutations and combinations (counting methods, factorial notation, ⁿPᵣ and ⁿCᵣ) are included in FBISE Class 11 Mathematics and O-Level Additional Mathematics. IB students cover these topics extensively in both AA and AI Higher Level courses.

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Permutations & Combinations Explained