Combinations and permutations
Work out nCr and nPr — how many ways to choose r items from n, with and without order.
C(n, r) = n! / (r! x (n - r)!) P(n, r) = n! / (n - r)! P(52, 5) = C(52, 5) x 5!
n! = 80 658 175 170 943 878 571 660 636 856 403 766 975 289 505 440 883 277 824 000 000 000 000
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Combinations and permutations: the only difference is whether order counts
Both answer “how many ways can I pick r things from n?”, and they differ by exactly one factor: r!. If the order of your picks matters you want permutations; if it does not, you want combinations, which is the smaller number because every ordering of the same group collapses into one.
How it works
- Computes nCr and nPr exactly, including for values far past what a normal number type can hold.
- Shows the factorials behind the result, so the formula is checkable rather than opaque.
- Includes presets for the two cases people actually arrive with: lottery draws and card hands.
C(n, r) = n! / (r! x (n - r)!) P(n, r) = n! / (n - r)! P(n, r) = C(n, r) x r! ordered arrangements = unordered groups x the ways to order each group
Worked example
Five-card poker hands from a 52-card deck.
- n = 52, r = 5
- P(52,5) = 52 x 51 x 50 x 49 x 48 = 311,875,200
- 5! = 120
- C(52,5) = 311,875,200 / 120 = 2,598,960
2,598,960 distinct hands. The permutation count is 120 times larger because it counts the same five cards dealt in each of 5! = 120 orders as different outcomes — which matters for a sequence lock and does not for a poker hand.
Reading the result
- nCr is symmetric: choosing 5 from 52 has the same count as choosing 47, because picking a group is the same act as picking everyone left out. This tool exploits that, always looping over the smaller of the two so the arithmetic stays cheap.
- Results are computed as exact integers, not floating point. 21! already exceeds what a double can represent exactly, and 100C50 is a 30-digit number — a calculator using ordinary numbers starts returning approximations well before that and rarely says so.
- Computing n!/(r!(n−r)!) literally overflows almost immediately. This divides at each step instead, so the running value never grows beyond the answer itself.
Common questions
- Which one do I want for a lottery?
- Combinations. A draw of 6 from 49 does not care what order the balls come out in, so it is C(49,6) = 13,983,816. That is the count of distinct tickets, and one in that many is your chance with a single line.
- Why is nPr always bigger?
- Because every unordered group is counted r! times over, once for each way of arranging it. With r = 5 that is 120 orderings of the same five items, so the permutation count is exactly 120 times the combination count.