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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.

Cập nhật Ví dụ đã kiểm tra: 9

Thử
Number of ways
Number of ways: 2,598,960
Số nguyên; Gần nhất; nếu cách đều thì ra xa số không
Digits in the answer
7
Chance of one particular outcome
3.848 × 10⁻⁷

There are 2,598,960 ways to choose 5 of 52 items when order doesn't matter. If each is equally likely, any single one has a 3.848 × 10⁻⁷ chance — 1 in 2,598,960.

Size of each count for n = 52, r = 5 (number of digits, log₁₀ scale)

Combinations2,598,960Combinations (repeats)3,819,816Permutations311,875,200Permutations (repeats)380,204,032
Cách tính S

  1. Combinations

    (nr)=n!r! (n−r)!=52!5! 47!\binom{n}{r} = \frac{n!}{r!\,(n-r)!} = \frac{52!}{5!\,47!}

  2. Exact integer result

    =2,598,960= 2{,}598{,}960

    Computed in exact BigInt arithmetic.

  3. Chance of one particular outcome

    12,598,960=3.84769×10−7\frac{1}{2{,}598{,}960} = 3.84769 \times 10^{-7}

    Assumes every outcome is equally likely.

Giới thiệu Permutation and combination calculator (nPr, nCr)

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.

Ví dụ có lời giải

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

Nguồn đối chiếu: 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

Nguồn đối chiếu: 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

Nguồn đối chiếu: 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

Nguồn đối chiếu: Python math.comb(100, 50)

Câu hỏi

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.

“Permutation and combination calculator (nPr, nCr)” chính xác đến mức nào?

Độ chính xác phụ thuộc vào dữ liệu nhập và giả định của phương pháp. Phép tính thập phân dùng 50 chữ số có nghĩa, nhưng ước lượng, phương pháp số và dữ liệu nguồn có thể kém chính xác hơn; làm tròn khi hiển thị không loại bỏ các giới hạn đó. Ví dụ có lời giải đã đối chiếu với nguồn độc lập: 9. Ví dụ, “5-card hands from 52 (defaults)” được kiểm tra bằng C(52,5) = 2,598,960 (Python math.comb; the standard count of poker hands).

Phương pháp này lấy từ đâu?

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).

Về công cụ tính này

(nr)=n!r!(n−r)!,P(n,r)=n!(n−r)!,(n+r−1r),nr\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

Nguồn

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

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