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Math

Permutations & Combinations (nPr / nCr) Calculator

Enter the total number of items n and how many to pick r to instantly get the permutations P(n, r), combinations C(n, r) and the factorial n!. Common uses include Mark Six / lottery odds, counting poker hands, sizing brute-force search spaces, and discrete-math / probability homework.

Examples: 49 choose 6 (Mark Six / Lotto), 52 choose 5 (poker hands)

Combinations C(n, r)

13,983,816

Order does not matter ({a,b,c} = {c,b,a}). Also written nCr or "n choose r".

Permutations P(n, r)

10,068,347,520

Order matters: {a,b,c} ≠ {c,b,a}. Also written nPr.

n! (n factorial)

608,281,864,034,267,560,872,252,163,321,295,376,887,552,831,379,210,240,000,000,000

n × (n−1) × … × 1. 0! = 1 by convention.

Formula

C(n, r) = n! ÷ (r! × (n − r)!) // order does not matter — also written nCr or "n choose r" P(n, r) = n! ÷ (n − r)! // order matters — also written nPr n! = n × (n − 1) × … × 1 // 0! = 1 by convention Probability of one specific combination at random = 1 ÷ C(n, r)

Frequently asked

What's the difference between permutations and combinations?

In short: permutations care about order, combinations don't. Picking 3 people out of 10 to go to lunch — only the set matters — is C(10, 3) = 120. Picking 3 people for distinct roles (chair / vice-chair / secretary) is a permutation P(10, 3) = 720 because role assignments differ. They're related by P(n, r) = C(n, r) × r!.

Why are the Mark Six / lottery odds 1 in 13,983,816?

Hong Kong Mark Six draws 6 numbers from 1–49 and the order doesn't matter — only the set of 6 needs to match — so there are C(49, 6) = 13,983,816 possible draws. One ticket (one chosen set of six) wins the jackpot with probability 1 ÷ 13,983,816 ≈ 7.15 × 10⁻⁸. Type n = 49, r = 6 to see this directly.

Why is 0! defined to be 1?

0! = 1 is a convention chosen so that identities stay clean. It makes C(n, 0) = n! ÷ (0! × n!) = 1 (there's exactly one way to pick "nothing" — the empty set) and the boundary terms of the binomial expansion (a + b)ⁿ work without special-casing. Setting 0! = 0 would put a zero in the denominator of every combinatorial formula. Combinatorially, 0! counts the empty permutation, which has exactly one instance.

How many 5-card poker hands are possible from a 52-card deck?

Choosing 5 cards out of 52, ignoring order, gives C(52, 5) = 2,598,960 distinct hands — enter n = 52, r = 5 to verify. The odds of being dealt a royal flush (only 4 possible — one per suit): 4 ÷ 2,598,960 ≈ 1 in 649,740.

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