What is 52 choose 5?
52C5 = 2,598,960 distinct five-card hands from a standard 52-card deck.
10C3 = 120
Count unordered selections nCr—poker hands, committees, and lottery lines where {A,B,C} equals {C,B,A}.
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nCr counts unordered selections: how many ways to choose r items from n when order does not matter. 52C5 = 2,598,960 distinct five-card poker hands from a standard deck.
10C3 = 120
Combinations count groups, not sequences. A committee of three from ten people has the same members whether Alice is listed first or third—only the set {Alice, Bob, Carol} matters.
Formula: nCr = n! / (r! × (n−r)!) — often read “n choose r.” The extra r! in the denominator strips the r! ways to reorder each subset that permutations would treat as different.
**Relationship:** nCr = nPr / r!. If 5P3 = 60 ordered triples, dividing by 3! = 6 gives 5C3 = 10 unordered triples.
Pascal's triangle rows list binomial coefficients C(n,k). Row n sums to 2^n—the number of subsets of an n-element set. Combinations appear in probability whenever drawing without order: lottery balls, committees, card hands.
Order ignored → nCr. Enter n and r for exact “n choose r,” then use probability calculator for event fractions.
Concise answers for common searches — definitions, steps, and comparisons.
52C5 = 2,598,960 distinct five-card hands from a standard 52-card deck.
Divide by r!: nCr = nPr / r!. Example: 5P3/3! = 60/6 = 10 = 5C3.
10C3 = 120 unordered triples from ten distinct items.
There is exactly one way to choose nothing—the empty subset.
Computes nCr via exact factorial ratios with BigInt cancellation, avoiding overflow from computing n! alone.
Formula
nCr = n! / (r! (n−r)!) for 0 ≤ r ≤ n. Pascal rule: C(n,k) = C(n−1,k−1) + C(n−1,k). Binomial theorem: (x+y)^n = Σ C(n,k) x^k y^(n−k).If listing {1,2,3} is the same as {3,2,1}, combinations apply. Ranked awards need permutations.
Pool size n; subset size r with 0 ≤ r ≤ n.
nC(n−r) equals nCr—choosing 2 to leave out of 10 matches choosing 8 to keep.
Favorable combinations ÷ total nCr gives probability when outcomes are equally likely.
Input
n = 5, r = 2Output
5C2 = 10Ten unordered pairs from five elements.
Input
n = 10, r = 3Output
10C3 = 120120 teams of three—order of names on the roster ignored.
Input
n = 20, r = 18Output
20C18 = 20C2 = 190Choosing 18 to include equals choosing 2 to exclude.
Input
n = 52, r = 5Output
52C5 = 2,598,960Standard five-card hand count from full deck.
Input
n = 7, r = 0 and r = 7Output
7C0 = 7C7 = 1Empty set and full set each have exactly one combination.
Input
5P3 = 60Output
5C3 = 60/6 = 10Divide ordered count by 3! to drop order.
Common real-world scenarios where this tool saves time.
Five cards from 52: 52C5 = 2,598,960 possible hands—denominator for flush probabilities.
Pick 6 numbers from 49: 49C6 ≈ 13.98 million combinations—odds math starts here.
Choose 4 finalists from 20 applicants where rank within the group is irrelevant: 20C4.
Inspect 5 units from a batch of 100 without caring about inspection order: 100C5 sample sets.
Step-by-step chains that connect related tools for common tasks.
| n \ r | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| 4 | 1 | 4 | 6 | 4 | 1 |
| 5 | 1 | 5 | 10 | 10 | 5 |
| 6 | 1 | 6 | 15 | 20 | 15 |
| 7 | 1 | 7 | 21 | 35 | 35 |
Pascal's triangle rows—each entry sums to 2^n.
| Pool | Pick | nCr | Order of magnitude |
|---|---|---|---|
| 49 | 6 | 13,983,816 | ~14 million |
| 59 | 5 | 5,006,386 | ~5 million |
| 90 | 5 | 57,958,470 | ~58 million |
Compute 100C3 as (100×99×98)/(3×2×1) to avoid 100! entirely.
Row 6: 1,6,15,20,15,6,1—quick lookup for n ≤ 10 on exams.
Once favorable and total combinations are known, probability is their ratio.
Combinations never exceed permutations for the same n,r—catches mode errors.
Divide nPr by r! or use nCr directly—otherwise you overcount by every reordering.
nCr already counts each subset once. Do not divide by 2 again for pairs unless problem defines labeled ends.
Choosing 3 fruits from {apple, banana, cherry} allowing duplicates uses stars-and-bars, not nCr.
Standard 52C5 treats cards as distinct—even same rank different suits are different items.
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10C3 = 120 ignores order. 10P3 = 720 counts every ordering—720/6 = 120.
Yes—use symmetry nCr = nC(n−r) to multiply fewer terms.
Exactly k successes in n trials has C(n,k) × p^k × (1−p)^(n−k) favorable patterns.
No—roles are ordered. Use permutations or multiply combinations by r! for labeled positions.
Also called multiset combinations—formula C(n+r−1, r), not standard nCr.
n!/(n−r)! is nPr; dividing by r! removes order. Together they count subsets only.
Peaks near r = n/2. 100C50 has 30 digits—use exact tools, not floating calculators.
Combination inputs are calculated locally without server upload.
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Reviewed by EverydayTools Editorial Team on 2026-07-03.
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