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Калькулятор перестановок и сочетаний (nPr / nCr)

ДанныеРазработчик

ВХОД

Автоматический процесс

ВЫХОД

Клиентская сторона

Результаты

Свойство	Ценить
P(n, r) — Permutations	-
C(n, r) — Combinations	-
n!	-
r!	-
(n - r)!	-
Multiset Permutations	-

Step-by-Step Formulas

Pascal's Triangle

Enumeration

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Гид

Permutation & Combination Calculator

Calculate permutations (nPr) and combinations (nCr) with exact results using BigInt arithmetic — no overflow even for large numbers. Enter n and r to get both values simultaneously with step-by-step formula breakdowns. Includes a Pascal’s Triangle visualizer, multiset permutation calculator, and full enumeration of all arrangements for small sets.

Как использовать

Enter n (total items) and r (items to choose) and the calculator instantly shows P(n,r) and C(n,r) with the formulas and step-by-step substitution. For small values of n and r (up to 8), toggle the enumeration to see all actual permutations or combinations listed out. Use the Pascal’s Triangle tab to visualize binomial coefficients up to row 20, with your C(n,r) value highlighted. The multiset tab handles permutations with repeated elements.

Возможности

nPr & nCr – Calculate both permutations and combinations simultaneously from a single input
BigInt Arithmetic – Exact results for large numbers (n=100+) with no overflow or rounding errors
Step-by-Step Formulas – See the formula with actual numbers substituted: P(n,r) = n!/(n-r)! and C(n,r) = n!/(r!×(n-r)!)
Pascal’s Triangle – Interactive visualizer showing rows 0 through n (up to 20) with your result highlighted
Full Enumeration – List all actual permutations or combinations for small sets (n,r ≤ 8)
Multiset Permutations – Calculate arrangements with repeated elements: n!/(n1!×n2!×…)
Factorial Display – Shows n!, r!, and (n-r)! values
Копировать результаты – Копировать все рассчитанные значения в буфер обмена

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 Часто задаваемые вопросы

What is the difference between permutations and combinations?

Permutations count arrangements where order matters. Combinations count selections where order does not matter. For example, choosing 3 letters from ABC: the permutations include ABC, ACB, BAC, BCA, CAB, CBA (6 arrangements), while the combination is just one set {A,B,C}. The formula for permutations P(n,r) = n!/(n-r)! gives a larger result than combinations C(n,r) = n!/(r!(n-r)!) because permutations count each ordering separately. Use permutations for rankings, sequences, and codes. Use combinations for teams, groups, and selections.
Why does this calculator use BigInt instead of regular numbers?

JavaScript regular numbers lose precision beyond 2^53 (about 9 quadrillion). Factorials grow extremely fast: 20! is already over 2.4 quintillion, and 100! has 158 digits. Regular floating-point arithmetic would give rounded or incorrect results for any moderately large calculation. BigInt provides exact integer arithmetic with no upper limit, ensuring every digit of your result is correct. This matters when you need precise counts for probability, statistics, or combinatorial analysis.
What is Pascal's Triangle and how does it relate to combinations?

Pascal's Triangle is a triangular array where each number is the sum of the two numbers directly above it. Row n of the triangle contains all values of C(n,r) for r from 0 to n. For example, row 4 is 1, 4, 6, 4, 1, which are C(4,0) through C(4,4). The triangle reveals patterns in binomial coefficients and has applications in probability, algebra, and number theory. Each row also gives the coefficients when expanding (a+b)^n. This calculator visualizes Pascal's Triangle and highlights your specific C(n,r) value.
What are multiset permutations?

Multiset permutations count the distinct arrangements of a collection that contains repeated elements. The formula is n!/(n1! times n2! times ... times nk!), where n is the total number of items and n1, n2, etc. are the counts of each repeated element. For example, the word MISSISSIPPI has 11 letters with M appearing 1 time, I appearing 4 times, S appearing 4 times, and P appearing 2 times. The number of distinct arrangements is 11!/(1! times 4! times 4! times 2!) = 34,650. Without accounting for repetition, you would overcount identical arrangements.