Introduction to Permutations

Learn how to count arrangements where order matters using permutations.

Advanced25 minLesson
Printable worksheet

Definition

A permutation is an arrangement of objects where order matters.
The number of ways to arrange nn distinct objects in a row is:
n!=n×(n−1)×(n−2)×⋯×2×1n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1
This is called n factorial.
When selecting and arranging rr objects from nn objects:
P(n,r)=n!(n−r)!=n×(n−1)×⋯×(n−r+1)P(n,r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)
Key insight: In permutations, ABC is different from BAC because the order is different.

Try it now

What is 3!3! (3 factorial)?

Worked Examples

In how many ways can 4 students (Anna, Ben, Carla, David) line up for a photo?
1

Identify what we're counting

We need to arrange all 4 students in a line where position matters → This is a permutation of 4 objects
2

Count choices for each position

1st position: 4 choices
2nd position: 3 remaining
3rd position: 2 remaining
4th position: 1 remaining → 4×3×2×14 \times 3 \times 2 \times 1
3

Calculate using factorial

4!=4×3×2×14! = 4 \times 3 \times 2 \times 1 → 4!=244! = 24
4

Interpret the result

There are 24 different ways to arrange 4 students → 24 arrangements

Common Mistakes

Confusing permutations and combinations

Why it's wrong: Permutations count arrangements where ORDER MATTERS. Combinations count selections where order doesn't matter.

Correct: Ask yourself: "Does rearranging change the outcome?" If selecting a committee (no positions), order doesn't matter = combination. If electing officers, order matters = permutation.

Using the wrong formula direction: (n−r)!n!\frac{(n-r)!}{n!} instead of n!(n−r)!\frac{n!}{(n-r)!}

Why it's wrong: The larger factorial n!n! goes on top because we start with more choices.

Correct: P(n,r)=n!(n−r)!P(n,r) = \frac{n!}{(n-r)!}, remember: n!n! is always larger, so it goes in the numerator.

Forgetting that 0!=10! = 1

Why it's wrong: When arranging all n objects, we use P(n,n)=n!0!=n!1=n!P(n,n) = \frac{n!}{0!} = \frac{n!}{1} = n!

Correct: By definition, 0!=10! = 1. This makes the formula work when r=nr = n.

Interactive Visual

Filling the places

4
×
3
×
2
×
1
=
24

The number under each place is how many choices are left for it.

4! = 4 × 3 × 2 × 1 = 24

Tap an item for the next place (4 choices).

Filling the places

5
×
4
×
3
=
60

The number under each place is how many choices are left for it.

P(5, 3) = 5!/(5 − 3)! = 5 × 4 × 3 = 60

Tap an item for the next place (5 choices).

Interactive Sandbox

Expression Calculator

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History

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Practice Problems

18 problems
Problem 1 of 18
Easy
What is 3!3! (3 factorial)?

Why It Matters

Permutations help us count arrangements in countless real-world situations:
  • Passwords: How many 4-digit PINs use different digits?
  • Races: How many ways can gold, silver, bronze medals be awarded?
  • Scheduling: In how many orders can 5 tasks be completed?
  • Seating: How many ways can students sit in a row?
Understanding permutations is essential for probability, cryptography, and computer science!

Real World Applications

Password Security

Understanding permutations helps us calculate password strength and security.

Example:

A 4-digit PIN using digits 0-9 without repetition has P(10,4)=5040P(10,4) = 5040 possibilities.

1Try It Yourself

A website requires a 3-character password using only uppercase letters (A-Z), with no letter repeated.

How many possible passwords are there?

Step 1: Write the mathematical expression

Calculate P(26,3)P(26,3):

Sports Tournament Brackets

Permutations determine possible outcomes in competitions where finishing position matters.

Example:

In a 6-team playoff, the possible final standings are 6!=7206! = 720 different orders.

2Try It Yourself

An Olympic swimming final has 8 swimmers. Medals are awarded for 1st, 2nd, and 3rd place.

How many different medal outcomes are possible?

Step 1: Write the mathematical expression

Use P(8,3)P(8,3):

Key Takeaways

  • 1A permutation is an arrangement where order matters
  • 2To arrange all nn objects: use n!=n×(n−1)×⋯×1n! = n \times (n-1) \times \cdots \times 1
  • 3To arrange rr objects from nn objects: use P(n,r)=n!(n−r)!P(n,r) = \frac{n!}{(n-r)!}
  • 4Quick method: multiply n×(n−1)×⋯n \times (n-1) \times \cdots for rr terms
  • 5Remember: 0!=10! = 1 by definition

Frequently Asked Questions

Ask: "Does the order matter?" If electing a President AND Vice President = permutation (order matters). If choosing 2 people for a committee = combination (order doesn't matter).
Ask: "Does the order matter?" If electing a President AND Vice President = permutation (order matters). If choosing 2 people for a committee = combination (order doesn't matter).
By convention, 0!=10! = 1 because there's exactly ONE way to arrange zero objects (do nothing). It also makes formulas like P(n,n)=n!0!=n!P(n,n) = \frac{n!}{0!} = n! work correctly.
P(n,r) means: "The number of ways to select AND arrange r objects from n objects." For example, P(5,3)P(5,3) = ways to choose and order 3 items from 5.

Glossary

Permutation
An arrangement of objects where order matters
Factorial
The product of all positive integers up to n, written as n!n!
P(n,r)
The number of permutations of r objects selected from n objects
Distinct objects
Objects that are all different from each other

Formula Card

Factorial

n!=n×(n−1)×(n−2)×⋯×2×1n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

Product of all positive integers up to n

Permutation

P(n,r)=n!(n−r)!P(n,r) = \frac{n!}{(n-r)!}

Arrangements of r objects from n

Special cases

0!=10! = 1, 1!=11! = 1, P(n,n)=n!P(n,n) = n!

Important values to remember

Today's challengeOne problem, three tries

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