Permutation and Combination Calculator
A permutation and combination calculator is a mathematical tool that computes the number of ways to arrange or select items from a set, using factorial notation and combinatorial formulas to solve nPr (permutations where order matters) and nCr (combinations where order doesn't matter) problems. These fundamental counting principles enable students, statisticians, probability theorists, and researchers to calculate arrangements, selections, lottery odds, sample spaces, binomial coefficients, and probability distributions for discrete mathematics, statistics courses, competitive exams, cryptography, genetics, quality control sampling, and any application requiring systematic counting of possible outcomes in ordered or unordered selections.
🔢 Interactive Permutation & Combination Calculator
Calculate nPr, nCr, and Factorials
Permutation Calculator (nPr)
Calculate arrangements where order matters
Combination Calculator (nCr)
Calculate selections where order doesn't matter
Factorial Calculator (n!)
Calculate factorial of a number
Understanding Permutations and Combinations
Permutations and combinations are fundamental concepts in combinatorics that count the number of ways to arrange or select items. The key difference is whether the order of selection matters.
Permutation Formula
Definition and Formula
Permutation Formula (nPr):
\[ P(n,r) = \frac{n!}{(n-r)!} \]
Where:
\( n \) = Total number of items
\( r \) = Number of items to arrange
\( n! \) = n factorial = \( n \times (n-1) \times (n-2) \times ... \times 2 \times 1 \)
Notation: \( nPr \) or \( P(n,r) \) or \( ^nP_r \)
Permutation Example
Problem: How many ways can you arrange 3 books from a shelf of 5 books?
Given: n = 5, r = 3
Formula: \( P(5,3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} \)
Step 1: Calculate factorials
5! = 5 × 4 × 3 × 2 × 1 = 120
2! = 2 × 1 = 2
Step 2: Apply formula
\( P(5,3) = \frac{120}{2} = 60 \)
Answer: There are 60 different ways to arrange 3 books from 5 books.
Combination Formula
Definition and Formula
Combination Formula (nCr):
\[ C(n,r) = \frac{n!}{r!(n-r)!} \]
Where:
\( n \) = Total number of items
\( r \) = Number of items to select
\( n! \) = n factorial
Notation: \( nCr \) or \( C(n,r) \) or \( \binom{n}{r} \) or \( ^nC_r \)
Combination Example
Problem: How many ways can you select 3 students from a class of 10?
Given: n = 10, r = 3
Formula: \( C(10,3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3! \times 7!} \)
Step 1: Simplify
\( C(10,3) = \frac{10 \times 9 \times 8 \times 7!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3!} \)
Step 2: Calculate
\( C(10,3) = \frac{720}{6} = 120 \)
Answer: There are 120 different ways to select 3 students from 10.
Key Differences: Permutation vs Combination
| Aspect | Permutation (nPr) | Combination (nCr) |
|---|---|---|
| Order | Order matters | Order doesn't matter |
| Formula | \( \frac{n!}{(n-r)!} \) | \( \frac{n!}{r!(n-r)!} \) |
| Result | Arrangements | Selections |
| Example | Seating arrangements | Team selection |
| Value | Always ≥ nCr | Always ≤ nPr |
| Keywords | "arrange", "order", "rank" | "choose", "select", "group" |
Factorial Definition
Factorial (n!):
\[ n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1 \]
Special Cases:
\( 0! = 1 \) (by definition)
\( 1! = 1 \)
Examples:
3! = 3 × 2 × 1 = 6
5! = 5 × 4 × 3 × 2 × 1 = 120
10! = 3,628,800
Factorial Reference Table
| n | n! | n | n! |
|---|---|---|---|
| 0 | 1 | 8 | 40,320 |
| 1 | 1 | 9 | 362,880 |
| 2 | 2 | 10 | 3,628,800 |
| 3 | 6 | 11 | 39,916,800 |
| 4 | 24 | 12 | 479,001,600 |
| 5 | 120 | 15 | 1,307,674,368,000 |
| 6 | 720 | 20 | 2.43 × 10¹⁸ |
| 7 | 5,040 | 25 | 1.55 × 10²⁵ |
Relationship Between nPr and nCr
Relationship Formula:
\[ nPr = nCr \times r! \]
Or equivalently:
\[ nCr = \frac{nPr}{r!} \]
This shows that permutations are combinations multiplied by all possible arrangements (r!) of selected items.
Common Formulas and Properties
Combination Properties
| Property | Formula |
|---|---|
| Symmetry | \( C(n,r) = C(n,n-r) \) |
| Choosing None | \( C(n,0) = 1 \) |
| Choosing All | \( C(n,n) = 1 \) |
| Pascal's Identity | \( C(n,r) = C(n-1,r-1) + C(n-1,r) \) |
| Sum Property | \( \sum_{r=0}^{n} C(n,r) = 2^n \) |
Special Cases
Permutation of n objects all taken: \( P(n,n) = n! \)
Permutation with repetition: \( n^r \)
Circular permutation: \( (n-1)! \)
Permutation with identical objects: \( \frac{n!}{n_1! \times n_2! \times ... \times n_k!} \)
Real-World Applications
Permutation Applications
- Password creation: Number of possible passwords with ordered characters
- Race rankings: Different podium finishes (gold, silver, bronze)
- Seating arrangements: Ways to arrange people in specific seats
- Lock combinations: Ordered digit sequences
- Competition rankings: First, second, third place awards
- DNA sequences: Order of nucleotide bases
Combination Applications
- Lottery selections: Choosing winning numbers (order irrelevant)
- Committee formation: Selecting team members
- Card hands: Poker hands, bridge deals
- Menu choices: Selecting pizza toppings
- Sample selection: Quality control sampling
- Genetics: Possible gene combinations
Step-by-Step Problem Solving
How to Identify Permutation vs Combination
Decision Flowchart:
- Does order matter?
- YES → Use Permutation (nPr)
- NO → Use Combination (nCr)
- Key words for Permutation: arrange, order, rank, sequence, position
- Key words for Combination: choose, select, group, pick, committee
- Test question: If I swap two items, is it a different outcome?
- YES → Permutation
- NO → Combination
Sample Calculation Table
| n | r | nPr | nCr | Relationship |
|---|---|---|---|---|
| 5 | 2 | 20 | 10 | 20 = 10 × 2! |
| 5 | 3 | 60 | 10 | 60 = 10 × 3! |
| 6 | 2 | 30 | 15 | 30 = 15 × 2! |
| 10 | 3 | 720 | 120 | 720 = 120 × 3! |
| 8 | 4 | 1,680 | 70 | 1,680 = 70 × 4! |
Common Problem Types
Problem Type 1: Arranging Letters
Question: How many different 4-letter arrangements can be made from the word "COMPUTER"?
Solution: n = 8 letters, r = 4 positions
\( P(8,4) = \frac{8!}{(8-4)!} = \frac{8!}{4!} = \frac{40,320}{24} = 1,680 \)
Answer: 1,680 arrangements
Problem Type 2: Committee Selection
Question: A committee of 4 people is to be formed from 9 candidates. How many ways can this be done?
Solution: n = 9, r = 4 (order doesn't matter)
\( C(9,4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4! \times 5!} = \frac{362,880}{24 \times 120} = 126 \)
Answer: 126 ways
Problem Type 3: Probability Applications
Question: What is the probability of being dealt 2 aces from a standard deck of 52 cards?
Solution:
Favorable outcomes: C(4,2) = 6 ways to choose 2 aces from 4
Total outcomes: C(52,2) = 1,326 ways to choose any 2 cards
Probability = 6/1,326 = 1/221 ≈ 0.0045 or 0.45%
Important Constraints and Conditions
⚠️ Conditions to Remember
- n ≥ r: Cannot select more items than available
- n ≥ 0, r ≥ 0: Both must be non-negative integers
- n! grows rapidly: 20! is already over 2 quintillion
- nPr ≥ nCr: Permutations always greater than or equal to combinations
- Calculator limits: Factorials beyond 170! may cause overflow
- 0! = 1: This is a convention, not intuitive but essential
Tips for Solving Problems
Problem-Solving Strategy:
- Read carefully: Identify if order matters
- Identify n and r: Determine total items and selection size
- Choose formula: Permutation or combination
- Check constraints: Ensure n ≥ r
- Calculate step-by-step: Break down factorials
- Verify answer: Does result make logical sense?
- Consider special cases: Restrictions, identical items
Frequently Asked Questions
What is the difference between permutation and combination?
Permutations count arrangements where order matters (ABC ≠ BAC), using formula nPr = n!/(n-r)!. Combinations count selections where order doesn't matter (ABC = BAC = CAB), using formula nCr = n!/[r!(n-r)!]. Example: Arranging 3 books from 5 is permutation (60 ways); selecting 3 books from 5 is combination (10 ways). Test: If swapping items creates different outcome, use permutation.
How do I calculate nPr and nCr manually?
For nPr: Calculate n!, then divide by (n-r)!. Example: 5P3 = 5!/(5-3)! = 120/2 = 60. For nCr: Calculate n!, divide by r! and (n-r)!. Example: 5C3 = 5!/(3!×2!) = 120/(6×2) = 10. Simplify before calculating: Cancel common terms in numerator and denominator to avoid large numbers. Use calculator for factorials above 10.
When should I use permutation vs combination?
Use permutation when: arranging items in order (seating charts, rankings, passwords), position matters (race medals), or swapping items changes outcome. Use combination when: selecting groups (committees, teams), order irrelevant (lottery numbers, card hands), or swapping items doesn't change outcome. Keywords: "arrange/order" → permutation; "choose/select" → combination.
What does 0! equal and why?
0! = 1 by mathematical convention and necessity. Reasons: (1) Maintains consistency in formulas like nC0 = n!/[0!×n!] = 1 (one way to choose nothing). (2) Recursive definition: n! = n×(n-1)!, so 1! = 1×0!, thus 0! = 1. (3) Empty product: Product of no numbers is multiplicative identity = 1. This definition makes formulas work correctly for all cases.
How do I solve problems with restrictions?
Identify restrictions first: Must together (treat as single unit), must separate (subtract restricted cases), specific positions (fix those positions, arrange rest), or excluded items (reduce n). Example: Arrange 5 people where 2 must sit together: Treat pair as 1 unit, arrange 4 units = 4! = 24, then arrange pair internally = 2! = 2, total = 24×2 = 48 ways.
What is circular permutation?
Circular permutation arranges n items in circle where rotation doesn't create new arrangement. Formula: (n-1)! instead of n!. Example: Seating 5 people at round table = (5-1)! = 24 arrangements. Rationale: Fix one person's position to eliminate rotational duplicates. If reflections also identical (like necklace), divide by 2: (n-1)!/2. Used for round tables, necklaces, circular arrangements.
Key Takeaways
Permutations and combinations are fundamental counting principles that solve arrangement and selection problems. Understanding when order matters determines which formula to apply for accurate calculation of possible outcomes.
Essential principles to remember:
- Permutation (nPr): Order matters, formula = n!/(n-r)!
- Combination (nCr): Order doesn't matter, formula = n!/[r!(n-r)!]
- Relationship: nPr = nCr × r!
- Factorial: n! = n × (n-1) × ... × 2 × 1, with 0! = 1
- Keywords: "arrange/order" → permutation; "choose/select" → combination
- Always check: n ≥ r (cannot select more than available)
- nPr ≥ nCr for same n and r values
- Circular permutation: (n-1)! arrangements
- Pascal's Identity: C(n,r) = C(n-1,r-1) + C(n-1,r)
- Applications: Probability, lottery, passwords, committees, genetics
Getting Started: Use the interactive calculator at the top of this page to compute permutations, combinations, and factorials. Enter your values for n (total items) and r (items to select/arrange) to get instant results with step-by-step explanations. Practice with different values to build intuition for when each formula applies.