# Permutation and combination calculator (nPr, nCr)

> Calculate nCr and nPr exactly: combinations and permutations with or without repetition, for poker hands, lottery odds and PIN codes.

Interactive version: https://www.calcopenly.com/statistics/permutations-combinations-calculator
Subject: Statistics and probability calculators

Combinations count the ways to pick r items from n when order doesn't matter: C(n, r) = n!/(r!(n − r)!). Permutations count each ordering separately, P(n, r) = n!/(n − r)!, so P(n, r) = C(n, r) × r!. When an item can be picked more than once, ordered picks number nʳ and unordered picks C(n + r − 1, r), the "stars and bars" count. Every answer is an exact whole number, up to 20,000 digits.

The default, 5 cards from 52, gives 2,598,960 poker hands, so one particular hand has a 1 in 2,598,960 chance. The same arithmetic gives 13,983,816 tickets in a 6-of-49 lottery, 720 podium orders from 10 runners and 10,000 four-digit PINs.

The chance of one outcome assumes every outcome is equally likely. Answers of 22 digits or more are shown in scientific notation, with every digit listed up to 300 digits.

## Inputs

- **Count** (options: Combinations — order doesn't matter, Combinations with repetition, Permutations — order matters, Permutations with repetition)
- **Items to choose from (n)**
- **Items chosen (r)**

## Results

- Number of ways — main result
- Digits in the answer
- Every digit
- Chance of one particular outcome

## Formula

$$
\binom{n}{r} = \frac{n!}{r!(n-r)!},\quad P(n,r) = \frac{n!}{(n-r)!},\quad \binom{n+r-1}{r},\quad n^r
$$

## Worked examples

### 5-card hands from 52 (defaults)

- Count: Combinations — order doesn't matter
- Items to choose from (n): 52
- Items chosen (r): 5
- **Number of ways: 2,598,960**
- **Digits in the answer: 7**
- **Chance of one particular outcome: 3.848 × 10⁻⁷**
- Checked against: C(52,5) = 2,598,960 (Python math.comb; the standard count of poker hands)

### Podium from 10 runners

- Count: Permutations — order matters
- Items to choose from (n): 10
- Items chosen (r): 3
- **Number of ways: 720**
- Checked against: 10 × 9 × 8 = 720 (Python math.perm)

### 6 of 49 lottery

- Count: Combinations — order doesn't matter
- Items to choose from (n): 49
- Items chosen (r): 6
- **Number of ways: 13,983,816**
- **Chance of one particular outcome: 7.151 × 10⁻⁸**
- Checked against: Python math.comb(49, 6) = 13,983,816; 1/13983816 = 7.15112e-8

### C(100, 50) exactly

- Count: Combinations — order doesn't matter
- Items to choose from (n): 100
- Items chosen (r): 50
- **Number of ways: 1.00891 × 10²⁹**
- **Digits in the answer: 30**
- Checked against: Python math.comb(100, 50)

### 3 scoops from 5 flavours, repeats allowed

- Count: Combinations with repetition
- Items to choose from (n): 5
- Items chosen (r): 3
- **Number of ways: 35**
- Checked against: C(5 + 3 − 1, 3) = C(7, 3) = 35 (Python math.comb)

### 4-digit PIN codes

- Count: Permutations with repetition
- Items to choose from (n): 10
- Items chosen (r): 4
- **Number of ways: 10,000**
- Checked against: 10⁴ = 10,000: each of the 4 positions takes any of 10 digits (Python 10**4)

## Questions

### What is the difference between a permutation and a combination?

A permutation counts orderings; a combination counts only which items are chosen. From 10 runners there are P(10, 3) = 720 ways to fill gold, silver and bronze, but only C(10, 3) = 120 different groups of three, because each group can be ordered in 3! = 6 ways. Use permutations for rankings, seatings and passwords, and combinations for hands, committees and lottery tickets.

### How many 5-card poker hands are there?

There are C(52, 5) = 2,598,960 five-card hands from a standard 52-card deck, counting each set of cards once regardless of the order dealt. Only 4 of them are royal flushes, so the chance of being dealt one is 4 in 2,598,960, or 1 in 649,740. Counting ordered deals instead gives P(52, 5) = 311,875,200.

### How are lottery jackpot odds calculated?

Multiply the number of combinations for each drawn set. A 6-from-49 draw has C(49, 6) = 13,983,816 possible tickets. Powerball draws 5 of 69 white balls and 1 of 26 red balls, so the jackpot odds are C(69, 5) × 26 = 11,238,513 × 26, or 1 in 292,201,338. Order doesn't matter in these games, which is why combinations apply.

### How many 4-digit PIN combinations are there?

There are 10⁴ = 10,000 four-digit PINs, from 0000 to 9999, because each of the 4 positions can hold any of 10 digits. Strictly this is a permutation with repetition, since 1234 and 4321 are different codes; a "combination lock" is really a permutation lock. A 6-digit PIN has 10⁶ = 1,000,000 possibilities.

### Why is 0! equal to 1?

0! = 1 by definition, because there is exactly one way to arrange zero items: the empty arrangement. The convention keeps the formulas consistent, so C(n, 0) = n!/(0! × n!) = 1 and C(n, n) = 1. It also matches the gamma function, where 0! = Γ(1) = 1, and the recursion n! = n × (n − 1)! at n = 1.

### How accurate is the permutation and combination calculator?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 9 worked examples whose answers come from independent sources; for example, “5-card hands from 52 (defaults)” is checked against C(52,5) = 2,598,960 (Python math.comb; the standard count of poker hands).

### Where does the method come from?

NIST Digital Library of Mathematical Functions, §26.3 Lattice paths: binomial coefficients and §26.2 permutations; Graham, Knuth & Patashnik, Concrete Mathematics, 2nd ed., chapter 5 (binomial coefficients).

## Sources

- [NIST Digital Library of Mathematical Functions, §26.3 Lattice paths: binomial coefficients and §26.2 permutations](https://dlmf.nist.gov/26.3)
- Graham, Knuth & Patashnik, Concrete Mathematics, 2nd ed., chapter 5 (binomial coefficients)