Summary: This combination permutation calculator computes nCr, nPr and factorials with live steps, formulas and a chart. It accepts labeled numeric inputs, works offline through file:// and includes source-backed explanations for students, analysts and researchers.
Combination permutation calculator for nCr, nPr, repetition formulas and factorials using BigInt for large exact integers. The calculator works offline, updates instantly and includes a worked example, plain-text formula, MathML, references and structured data.
Default values are loaded. Click any field and edit it; results and chart update automatically.
| Input or setting | Result or interpretation | Use this when |
|---|---|---|
| 52 choose 5 | 2,598,960 | counting outcomes |
| 10P3 | 720 | counting outcomes |
| 0 factorial | 1 | counting outcomes |
P(n,r)=n!/(n-r)!; C(n,r)=n!/[r!(n-r)!]; with repetition: n^r or C(n+r-1,r)Permutations count ordered arrangements. Combinations count unordered selections.
n=52, r=5
nCr=2,598,960; nPr=311,875,200
This page uses the shared statistics core for distribution functions, quantiles and exact integer counting where needed. The formula is shown in plain text so screen readers and search engines can parse it reliably.
Use Load example in the calculator to reproduce this reference result.
{
"tool": "Combination Permutation Calculator",
"input": "n=52, r=5",
"output": "nCr=2,598,960; nPr=311,875,200",
"formula": "P(n,r)=n!/(n-r)!; C(n,r)=n!/[r!(n-r)!]; with repetition: n^r or C(n+r-1,r)"
}| Calculator | Example input | Expected output |
|---|---|---|
| Combination Permutation Calculator | n=52, r=5 | nCr=2,598,960; nPr=311,875,200 |
What does p = 0.03 mean? If the null hypothesis and model assumptions were true, a result at least this extreme would occur about 3% of the time. The American Statistical Association cautions that a p-value alone does not measure effect size, practical importance or the probability that Hâ‚€ is true.3
For most classroom and professional reports, pair the calculator result with the question you are answering. A mean or median summarizes location, but spread explains consistency. A confidence interval estimates plausible values, while a hypothesis test evaluates compatibility with a null model. Regression and correlation describe association, so they should be reported with a chart and residual or outlier review. When a result is statistically significant, still ask whether the effect is large enough to matter in the real setting.
| Statistic | Small | Medium | Large | Use |
|---|---|---|---|---|
| Cohen's d | 0.2 | 0.5 | 0.8 | t-test effect size |
| Cramér's V | 0.1 | 0.3 | 0.5 | chi-square association |
| |r| | 0.10 | 0.30 | 0.50 | correlation strength |
| R² | 0.01 | 0.09 | 0.25 | variance explained |
Simple rule — does order matter? If yes, it's a permutation. If no, combination. Picking gold, silver, and bronze medallists from 8 runners is a permutation, because finishing first is different from finishing third. Choosing 3 students from 8 to form a study group is a combination, because the group itself is the same regardless of who you pick first. Whenever you see words like "arrange", "sequence", or "rank", reach for nPr. For words like "select", "choose", or "pick", go with nCr.
Yes, that's the whole distinction. In permutations, ABC and BAC are counted as two different outcomes. In combinations, ABC and BAC are the same — they contain the same three letters, just rearranged. So nPr counts arrangements while nCr counts groups. A handy way to remember: combinations are like a salad bowl (mix doesn't matter), permutations are like a recipe (order changes the dish). You'll always get nPr ≥ nCr for the same n and r, because each combination produces r! permutations.
Don't compute the full factorials — they explode quickly. Instead, use nCr = n! / [r! (n−r)!] and cancel terms first. For C(100, 3), write it as (100 × 99 × 98) / (3 × 2 × 1) = 161700. You only need r terms in the numerator and r! in the denominator. Cancelling early keeps the numbers manageable. Also, nCr = nC(n−r), so C(100, 97) is the same as C(100, 3) — always pick the smaller r to make life easier.
When repetition is allowed, every position can be filled by any of the n items independently, so the count becomes nr — that's n multiplied by itself r times. A 4-character password using 26 lowercase letters allows 264 = 456,976 possibilities. A 4-digit PIN with digits 0–9 gives 104 = 10,000. Without repetition, the formula shrinks to nPr = n!/(n−r)!, because each choice reduces the pool. So for ordered slots with replacement, think of it as r independent decisions, each with n options.
The formula is C(n + r − 1, r), often called the "stars and bars" method. Suppose you're picking 3 scoops of ice cream from 5 flavours and repeats are allowed. That's C(5 + 3 − 1, 3) = C(7, 3) = 35 ways. The trick: imagine 3 stars (your scoops) and 4 bars separating the 5 flavour categories. Counting arrangements of these symbols gives the answer. This applies whenever you're selecting items with replacement and order doesn't matter — like choosing toppings or filling a multi-set.
It depends on whether you allow repeated digits. With repetition (the usual ATM case), each of the 4 slots has 10 choices, giving 104 = 10,000 possible PINs from 0000 to 9999. Without repetition, you'd use P(10, 4) = 10 × 9 × 8 × 7 = 5,040 — fewer options because once a digit's used, it's off the table. Most real PIN systems allow repeats, which is why people can pick 1111 (and please don't, that's the worst PIN you can choose).
This is a combination problem because the order in which you pick people doesn't change who's on the committee. A 5-person committee from 12 candidates gives C(12, 5) = 792 possible groups. The formula is C(n, r) = n! / [r!(n−r)!]. If the committee has distinct roles — say chair, secretary, treasurer — then order matters and it becomes a permutation: P(12, 3) = 1,320. So the question to ask first is whether all members are interchangeable or each has a labelled position.
nPr counts ordered arrangements; nCr counts unordered selections. The formulas are nPr = n!/(n−r)! and nCr = n!/[r!(n−r)!]. Notice nCr divides nPr by r! — that extra division removes the duplicate orderings. Picking 2 letters from {A, B, C}: permutations give AB, BA, AC, CA, BC, CB (six outcomes); combinations give just AB, AC, BC (three). So nPr is always larger than or equal to nCr for the same inputs, and they're equal only when r = 0 or r = 1.