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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.

آخر تحديث أمثلة تم التحقق منها: 9

جرّب
Number of ways
Number of ways: 2,598,960
عدد صحيح؛ إلى الأقرب، وعند التعادل بعيدًا عن الصفر
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
طريقة الحساب 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.

حول 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.

أمثلة محلولة

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

مصدر التحقق: ⁨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

مصدر التحقق: ⁨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

مصدر التحقق: ⁨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

مصدر التحقق: ⁨Python math.comb(100, 50)⁩

الأسئلة

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.

ما مدى دقة «⁨Permutation and combination calculator (nPr, nCr)⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 9. مثلًا، يجري التحقق من «⁨5-card hands from 52 (defaults)⁩» بالرجوع إلى ⁨C(52,5) = 2,598,960 (Python math.comb; the standard count of poker hands)⁩.

ما مصدر هذه الطريقة؟

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).

حول هذه الحاسبة

(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

المصادر

  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)

تم التحقق بالرجوع إلى المصادر

تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 9. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.

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