Calculateur de Permutation et de Combinaison (nPr / nCr)

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.

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.

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.

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.