# 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.

Versione interattiva: https://www.calcopenly.com/it/statistics/permutations-combinations-calculator
Argomento: Calcolatori di statistica e probabilità

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.

## Dati

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

## Risultati

- Number of ways — risultato principale
- Digits in the answer
- Every digit
- Chance of one particular outcome

## Formula

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

## Esempi svolti

### 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**
- Fonte di verifica: 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**
- Fonte di verifica: 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**
- Fonte di verifica: 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**
- Fonte di verifica: 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**
- Fonte di verifica: 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**
- Fonte di verifica: 10⁴ = 10,000: each of the 4 positions takes any of 10 digits (Python 10**4)

## Domande

### 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.

### Quanto è preciso «Permutation and combination calculator (nPr, nCr)»?

La precisione dipende dai dati inseriti e dalle ipotesi del metodo. Il calcolo decimale usa 50 cifre significative, ma stime, metodi numerici e dati di origine possono essere meno precisi; l’arrotondamento visualizzato non elimina questi limiti. Esempi svolti verificati con fonti indipendenti: 9. Per esempio, «5-card hands from 52 (defaults)» viene verificato con C(52,5) = 2,598,960 (Python math.comb; the standard count of poker hands).

### Da dove proviene il metodo?

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).

## Fonti

- [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)