Permutations and combinations answer the two counting questions that turn up everywhere from seating plans to password design: in how many ways can things be arranged, and in how many ways can they simply be chosen? The difference between the two β whether order counts β changes the answer dramatically, and confusing them is the most common counting error in mathematics. This calculator computes both from your n (the total number of items) and r (how many are taken at a time), so you can see the ordered and unordered counts side by side and pick the one your problem actually asks for.
Quick Answer
When order matters, use the permutation formula P(n, r) = n! Γ· (n β r)!. When order does not matter, use the combination formula C(n, r) = n! Γ· (r! Γ (n β r)!). For the same n and r, the permutation count is always larger β exactly r! times larger.
Order: The One Idea That Separates the Two Formulas
A permutation counts arrangements. Choosing a president, a vice-president and a treasurer from 10 people is a permutation problem because the same three people in different roles are different outcomes. A combination counts selections. Choosing a three-person committee from the same 10 people is a combination problem because the committee {Ana, Ben, Cy} is the same committee however you list its members.
The formulas encode that distinction. P(n, r) = n!/(nβr)! works through the slots one at a time: n choices for the first position, nβ1 for the second, and so on, stopping after r positions. C(n, r) runs the same slot-filling and then divides by r! β the number of ways those r chosen items could have been ordered among themselves β collapsing all the orderings of each selection into a single counted outcome. The exclamation mark is the factorial: n! means n Γ (nβ1) Γ ... Γ 2 Γ 1, with 0! defined as 1. Both formulas assume each item is used at most once; arrangements with repetition allowed (like PIN codes, where digits may repeat) follow a different rule, n^r, and are not what these formulas compute.
How to Use the Permutation and Combination Calculator
- Count the total pool of distinct items β that is n.
- Decide how many are being arranged or chosen at once β that is r.
- Ask the decisive question: would swapping the order of two chosen items create a different outcome? If yes, you need the permutation; if no, the combination.
- Enter n and r into the calculator's fields.
- Read both results and use the one matching your answer to step 3.
- Sanity-check the relationship: the permutation result should equal the combination result multiplied by r! (for r = 3, exactly 6 times).
Worked Example: Eight Runners, Two Different Questions
Eight runners finish a race. Question one: in how many ways can the gold, silver and bronze medals be awarded? Medals are ranked, so order matters β this is P(8, 3) = 8! / (8 β 3)! = 8 Γ 7 Γ 6 = 336 possible podiums. Question two: in how many ways can three of the eight be selected for a team photo, with no ranking? Now order is irrelevant β this is C(8, 3) = 8! / (3! Γ 5!) = (8 Γ 7 Γ 6) / (3 Γ 2 Γ 1) = 336 / 6 = 56 possible trios. Same eight people, same three chosen, and the answers differ by exactly 3! = 6, because each unordered trio can stand on the podium in six different orders. That factor-of-six check works for every pair of results this calculator gives you.
More reference values:
- P(5, 2) = 20, while C(5, 2) = 10
- P(10, 3) = 720, while C(10, 3) = 120
- P(6, 6) = 720 (arranging all six), while C(6, 6) = 1 (choosing all six)
- P(9, 4) = 3,024, while C(9, 4) = 126
Choosing the Wrong Formula, and Other Hazards
Nearly every error with these tools is conceptual rather than arithmetical. Awarding prizes, ranking finalists, assigning seats or posts, and forming sequences are permutation territory; committees, hands of cards, lottery selections, teams and samples are combination territory. If you cannot decide, ask whether the outcome "Ana first, Ben second" differs from "Ben first, Ana second" in the real situation β that sentence settles almost every case. Two technical hazards follow. First, r cannot exceed n in these formulas: you cannot choose 12 distinct people from a pool of 10, and a calculator fed r > n has nothing meaningful to return. Second, watch for hidden repetition: these formulas assume no item repeats, so a lock code of 4 digits from 10 is 10^4 = 10,000 (repetition allowed), not P(10, 4) = 5,040 (no repeats). Using the permutation formula there understates the count and any security conclusion drawn from it.
Handshakes and Round-Robins: Combinations in Disguise
Two party-and-sport questions turn out to be combinations wearing costumes. If everyone at a gathering of 12 people shakes hands with everyone else exactly once, how many handshakes happen? A handshake is an unordered pair β Ana-with-Ben is the same handshake as Ben-with-Ana β so the answer is C(12, 2) = (12 Γ 11) / 2 = 66 handshakes. A round-robin tournament is the identical problem with fixtures instead of hands: 10 teams, each playing each other once, need C(10, 2) = 45 matches, which is why league organisers reach for n(nβ1)/2 before they reach for a calendar. Recognising these as combinations matters because the wrong instinct is to multiply n by nβ1 and stop (132 "handshakes" for the party of 12), double-counting every pair. The formulas on this page are less about factorials than about this habit: deciding what one "outcome" is, and making sure each outcome is counted exactly once.
Where Counting Leads Next
If your problem is purely about selections, the Combination Calculator concentrates on that single formula with more selection-focused examples, and the Factorial Calculator exposes the building block both formulas share, which is handy when a textbook answer needs checking step by step. Counts like these are also the raw material of probability β the chance of a lottery combination or a particular hand starts life as a combination divided by a combination. The full range of counting, algebra and statistics tools is collected under Math.