Permutation Calculator

Understanding the two types of permutations.

Without Repetition (Standard)

P(n, r) = n! / (n-r)!

Or equivalently: P(n, r) = n × (n-1) × (n-2) × ... × (n-r+1)

Each item can only be used once.

Example: P(5, 3)

P(5, 3) = 5! / 2! = 120 / 2 = 60

Or: 5 × 4 × 3 = 60

With Repetition

P(n, r) = nʳ

Items can be reused in different positions.

Example: 4-digit PIN (digits 0-9)

P(10, 4) with repetition = 10⁴ = 10,000

Key Difference

Without: First position has n choices, second has n-1, etc. With: Every position has n choices.

The fundamental difference: does order matter?

Side-by-Side Comparison

The Relationship

P(n, r) = C(n, r) × r!

Permutations = Combinations × (ways to arrange each selection)

Example: 3 from 5

Combinations: C(5,3) = 10 groups {ABC}, {ABD}, {ABE}, {ACD}, {ACE}, {ADE}, {BCD}, {BCE}, {BDE}, {CDE}

Permutations: P(5,3) = 60 arrangements Each group of 3 can be arranged 3! = 6 ways 10 × 6 = 60

Choosing the Right One

Ask: "Does the order of selection matter?"

YES → Permutation:

• Ranking competitors
• Creating passwords
• Scheduling in sequence

NO → Combination:

• Selecting committee members
• Choosing lottery numbers
• Dealing card hands

Real-world problems using permutations.

Rankings and Contests

Olympic Medals (3 winners from 8): P(8, 3) = 8 × 7 × 6 = 336 possible podium outcomes

Class Elections (President, VP, Secretary from 25): P(25, 3) = 25 × 24 × 23 = 13,800

Security and Passwords

4-digit PIN (with repetition): P(10, 4) = 10⁴ = 10,000

6-character password (letters only, no repetition): P(26, 6) = 26 × 25 × 24 × 23 × 22 × 21 = 165,765,600

8-character password (letters + digits, with repetition): 36⁸ = 2,821,109,907,456

Scheduling

5 talks in 5 time slots: P(5, 5) = 5! = 120 possible schedules

3 meetings from 7 available slots: P(7, 3) = 7 × 6 × 5 = 210

Seating Arrangements

8 people in 8 chairs: 8! = 40,320 arrangements

5 people in a row of 12 seats: P(12, 5) = 95,040

Important permutation scenarios.

Arranging All Items

P(n, n) = n!

Arranging all n items uses all positions. P(5, 5) = 5! = 120

Circular Permutations

When arranged in a circle, one position is fixed: (n-1)! arrangements

Example: 5 people at a round table (5-1)! = 4! = 24 arrangements

Permutations with Identical Items

If some items are identical: n! / (n₁! × n₂! × ... × nₖ!)

Example: Arrangements of "MISSISSIPPI" 11! / (4! × 4! × 2!) = 34,650 (4 S's, 4 I's, 2 P's, 1 M)

Restricted Permutations

When certain positions have restrictions:

Example: 5 people, 2 must be adjacent Treat adjacent pair as one unit: 4! = 24 The pair can swap: × 2 Total: 48 arrangements

Derangements

Permutations where nothing is in its original position: D(n) ≈ n!/e

Example: No one gets their own hat D(4) = 9 arrangements

Fundamental rules that underlie permutations.

The Multiplication Principle

If task 1 can be done in m ways and task 2 in n ways: Total ways = m × n

This is why P(n,r) = n × (n-1) × ... × (n-r+1)

The Addition Principle

If events are mutually exclusive: Total = count₁ + count₂ + ...

Example: Arrangements starting with A OR B = (arrangements starting with A) + (arrangements starting with B)

Complement Counting

Sometimes easier to count what we DON'T want:

Desired = Total - Undesired

Example: Arrangements where A and B are NOT adjacent = Total arrangements - Arrangements where A and B ARE adjacent

Inclusion-Exclusion

For overlapping conditions: |A ∪ B| = |A| + |B| - |A ∩ B|

Useful when counting "at least one" scenarios.

Bijection Principle

If we can create a one-to-one correspondence between two sets, they have equal size.

Useful for proving permutation identities.

More complex counting scenarios.

Permutations with Forbidden Positions

Problem: Arrange 1,2,3,4 so no number is in its position

This is a derangement: D(4) = 4!(1 - 1/1! + 1/2! - 1/3! + 1/4!) = 9

Permutations with Required Adjacencies

Problem: 6 people, couple must sit together

Treat couple as one unit: 5! = 120 ways Couple can swap: × 2 Total: 240

Permutations with Forbidden Adjacencies

Problem: 5 books, math and physics can't be adjacent

Total arrangements: 5! = 120 Adjacent arrangements: 4! × 2 = 48 Non-adjacent: 120 - 48 = 72

Distribution to Distinct Boxes

n distinct items into k distinct boxes:

• All boxes get items: Stirling numbers × k!
• Any distribution: kⁿ

Permutations with Repetition Limits

Problem: 3-letter "words" from {A,B,C}, each used at most twice

Requires careful case analysis or generating functions.