How do permutations with replacement differ from without replacement?

Without replacement each item can only be chosen once, so nPr = n!/(n-r)!. With replacement items can repeat, so nPr = n^r. For combinations, without replacement gives C(n,r) = n!/(r!(n-r)!), while with replacement gives C(n+r-1, r).

Permutation & Combination Calculator

1

Enter n and r

Enter n (total items available) and r (items you are selecting or arranging) — for example, n = 10 players, r = 3 starting positions.

2

Choose Mode and Replacement

Select Permutation when order matters, Combination when it does not, then toggle With Replacement if items can repeat.

3

Review the Step-by-Step Work

Review the factorial expansion in the result card, then switch to the Scenarios tab for real-world examples or Pascal's Triangle to visualize C(n,r).

Permutation (without replacement)

nPr = n! / (n − r)!

Counts ordered arrangements of r items from n. Divide n! by (n − r)! to cancel the leftover items not selected.

Combination (without replacement)

nCr = n! / (r! × (n − r)!)

Counts unordered selections of r items from n. Divide by r! to eliminate the r! orderings of the chosen set that we no longer distinguish.

With Replacement

Permutation: nʳ | Combination: C(n+r−1, r)

When items can repeat: ordered arrangements grow exponentially (n choices per slot); unordered selections use the stars-and-bars formula C(n+r−1, r).

Permutation An ordered arrangement of r items chosen from n distinct items. ABC and BAC are different permutations because position matters.

Combination An unordered selection of r items from n. The sets {A,B,C} and {B,A,C} are the same combination because only membership matters, not order.

Factorial (n!) The product of all positive integers from 1 to n — for example, the factorial of 5 equals 5 × 4 × 3 × 2 × 1 = 120. By convention the factorial of zero equals 1, which keeps combination and.

n (population) The total number of distinct items available in the set from which you are selecting or arranging.

r (sample) The number of items selected or arranged. Must satisfy 0 ≤ r ≤ n for counting without replacement.

With vs Without Replacement Without replacement means each item appears at most once in the selection. With replacement means items can be chosen more than once, like rolling a die multiple times.

🏛️

Committee of 3 from 5 people

5 Choose 3

n 5 r 3 Result C(5,3) = 10

There are 10 different committees of 3 people from a group of 5. Only membership matters, not selection order, so this is a combination: C(5,3) = 5!/(3!×2!) = 120/12 = 10.

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1st and 2nd place from 6 racers

6P2 Race

n 6 r 2 Result P(6,2) = 30

There are 30 ways to assign 1st and 2nd place from 6 racers. Alice 1st, Bob 2nd differs from Bob 1st, Alice 2nd — order matters — so this is a permutation: P(6,2) = 6!/4! = 720/24 = 30.

🎱

Lottery — pick 6 numbers from 1–49

Lottery 6/49

n 49 r 6 Result C(49,6) = 13,983,816

Nearly 14 million possible tickets exist. Because drawn numbers win regardless of the order they were drawn, this is a combination. Your odds with one ticket are exactly 1 in 13,983,816.

Permutations and combinations answer the most fundamental questions in probability and combinatorics: in how many ways can things be arranged or selected? The two differ by one question — does order matter? Mastering this distinction unlocks probability calculations for card games, cryptography, statistics, and tournament design.

Passwords and Security

When you create an 8-character password from 94 printable ASCII characters with repetition allowed, there are 94⁸ ≈ 6.1 trillion possible passwords. This is a permutation with replacement — each of the 8 positions independently allows all 94 characters. Without repetition, you get P(94, 8) = 94 × 93 × 92 × … × 87 ≈ 5.5 trillion — only slightly fewer, because 8 characters out of 94 is a small fraction. Increasing password length by just one character multiplies possibilities by approximately 94×, which is why a 12-character password (94¹² ≈ 4.8 × 10²³) is astronomically more secure than an 8-character one, regardless of character variety. This is why security experts recommend length over complexity — adding one character to a 10-character password is more effective than switching from all-lowercase to mixed-case. Brute-force attack time grows exponentially with length; at one trillion guesses per second, cracking a 12-character password by exhaustive search would take billions of years. The combinatorics make 16+ character passphrases effectively uncrackable by any foreseeable hardware improvement.

Poker Hands

In a standard 52-card deck, the number of distinct 5-card hands is C(52, 5) = 2,598,960. Because a poker hand is a set of cards — not a sequence — order does not matter and combinations apply. Of those hands, exactly 4 are royal flushes (ace through ten of each suit), giving a probability of 4/2,598,960 ≈ 1 in 649,740. There are 36 total straight flushes (including royals), with probability ≈ 1 in 72,193. Full houses are counted as C(13,1) × C(4,3) × C(12,1) × C(4,2) = 3,744 hands, probability ≈ 1 in 694. Recognizing that poker hands are combinations — not permutations — is the essential first step in computing precise poker probabilities. Beginners sometimes overcount by treating the dealing order as meaningful, which would inflate hand counts by a factor of the 120 orderings of 5 cards, producing incorrect probability estimates. Two-pair hands, for instance, require careful double-counting avoidance: C(13,2) × C(4,2) × C(4,2) × C(44,1) = 123,552, not simply C(13,2)² × C(4,2)².

Tournament Brackets

In a single-elimination bracket of 64 teams, the number of possible outcomes is 2⁶³ ≈ 9.2 × 10¹⁸ — roughly 9.2 quintillion. Each of the 63 games has two possible winners, and game outcomes are ordered in the sense that which team wins each specific slot matters. This is most naturally framed as a product of independent binary choices: 2 × 2 × … × 2 (63 times) = 2⁶³. The astronomical size explains why a perfect bracket is essentially impossible even with expert knowledge: if you predict each game correctly with 70% probability — generous for later rounds — the chance of a perfect bracket is 0.7⁶³ ≈ 1 in 10¹², roughly one in a trillion. Warren Buffett famously offered $1 billion for a perfect NCAA bracket in Berkshire Hathaway promotional contests, a risk he could safely accept because the combinatorial odds make the prize effectively uncollectable. Even with domain knowledge reducing the number of genuinely uncertain games, the multiplication of independent uncertainties across 63 games overwhelms any prediction skill. This is why no perfect bracket has ever been submitted in any verified large-scale contest, despite millions of entries each year.

What is the difference between permutation and combination?+

A permutation counts ordered arrangements — ABC differs from BAC. A combination counts unordered selections — {A,B,C} equals {B,A,C}. A quick test: if swapping two items gives a different valid outcome, use a permutation; if the result is the same, use a combination.

What does 0! equal and why?+

By definition, the factorial of zero equals 1. This is required for the recursive relationship n factorial = n times (n-1) factorial to remain consistent at n = 1, and it ensures that C(n, 0) = 1, correctly reflecting that there is exactly one way to choose zero items from any set.

When do I use permutations versus combinations?+

Use permutations when sequence or position matters: ranking racers, generating PINs, scheduling tasks in order. Use combinations when only membership matters: forming committees, dealing cards, buying lottery tickets. Ask yourself: would rearranging the chosen items give a different valid outcome?

How do I handle very large n values?+

This calculator uses standard floating-point arithmetic, which supports n up to 170 before factorials overflow to infinity. For large n above about 20, the step-by-step display is simplified, but results remain accurate because n!/(n−r)! is computed by cancellation rather than computing full factorials.

What is the relationship between nPr and nCr?+

The relationship between permutations and combinations is nPr = nCr times r-factorial. Combinations count unordered groups; multiplying by the factorial of r reintroduces all the orderings, converting combinations to permutations.

Permutation & Combination Calculator

Inputs

Real-World Scenarios

Pascal's Triangle

Binomial Expansion

How to Use This Calculator

Enter n and r

Choose Mode and Replacement

Review the Step-by-Step Work

Formula & Methodology

Key Terms Explained

Real-World Examples

Committee of 3 from 5 people

1st and 2nd place from 6 racers

Lottery — pick 6 numbers from 1–49

Counting in the Real World

Passwords and Security

Poker Hands

Tournament Brackets

Frequently Asked Questions

Permutation & Combination Calculator

Inputs

Real-World Scenarios

Pascal's Triangle

Binomial Expansion

How to Use This Calculator

Enter n and r

Choose Mode and Replacement

Review the Step-by-Step Work

Formula & Methodology

Key Terms Explained

Real-World Examples

Counting in the Real World

Passwords and Security

Poker Hands

Tournament Brackets

Frequently Asked Questions

Keep Exploring

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