Combinations and permutations calculator guide
Combinatorics counts the ways to select or arrange items. A combination counts selections where order is irrelevant — choosing 3 people for a committee from 10 candidates is the same committee no matter the picking order. A permutation counts arrangements where order matters — awarding gold, silver, and bronze to 3 of 10 athletes gives a different outcome for every ordering.
Enter the total number of items n and the number chosen r, then select combination or permutation. The result is computed with exact big-integer arithmetic, so counts with hundreds of digits stay precise instead of rounding to scientific notation.
How to use this combinations and permutations calculator
- Decide whether order matters: Choose combination for unordered selections and permutation for ordered arrangements.
- Enter n: Enter the total number of distinct items as a whole number.
- Enter r: Enter how many items are chosen; r cannot exceed n.
- Calculate: Press Calculate for the exact count and the formula with your values filled in.
Formula and variables
A permutation arranges r items out of n: there are n choices for the first position, n − 1 for the second, and so on, giving n! ÷ (n − r)!. A combination divides the permutation count by the r! orderings of the same selection, because rearranging the chosen items does not create a new combination.
C(n, r) = n! ÷ (r! × (n − r)!) · P(n, r) = n! ÷ (n − r)!- n — Total items
- The size of the full set being chosen from. (count)
- r — Items chosen
- How many items are selected or arranged. (count)
- n! — Factorial
- The product n × (n − 1) × … × 2 × 1, with 0! = 1. (count)
- C(n, r) — Combinations
- Unordered selections, also written as the binomial coefficient “n choose r”. (count)
- P(n, r) — Permutations
- Ordered arrangements of r items out of n. (count)
Choosing a 3-person committee from 10 candidates
A club needs a 3-person committee; roles on the committee are identical, so order does not matter.
- Mode
- Combination
- Total items (n)
- 10
- Items to choose (r)
- 3
- C(10, 3) = 10! ÷ (3! × 7!)
- = (10 × 9 × 8) ÷ (3 × 2 × 1)
- = 720 ÷ 6
Result: 120 possible committees
If the three seats were distinct roles (president, secretary, treasurer), order would matter and the answer would be the permutation P(10, 3) = 720.
Understanding your results
Reading the count
The result is the exact number of distinct selections or arrangements. Permutations are always at least as large as combinations for the same n and r because each unordered selection of r items can be arranged in r! ways: P(n, r) = C(n, r) × r!.
When the numbers get large
Counts grow explosively — C(50, 5) is over two million and P(100, 10) has 20 digits. This calculator returns exact integers using big-integer arithmetic rather than floating-point approximations, so every digit shown is correct.
Assumptions
- Items are distinct and chosen without replacement; each item can be used at most once.
- n and r are non-negative whole numbers with r ≤ n.
- Combinations treat selections as unordered; permutations treat them as ordered.
Limitations
- Variations with repetition allowed (multichoose, or n^r arrangements) are not covered.
- Circular arrangements and identical-item (multiset) permutations use different formulas not included here.
- n is limited to 5,000 to keep the exact factorial computation responsive in the browser.
Common mistakes
- Using a permutation when the selection is unordered; this overcounts by a factor of r!.
- Using a combination when positions or roles are distinct, which undercounts.
- Forgetting that 0! = 1, so C(n, 0) = C(n, n) = 1.
- Entering r greater than n; you cannot choose more items than exist.
Practical use cases
Probability and statistics
Count favorable and total outcomes for lottery odds, card hands, and sampling problems — the binomial coefficient underlies the binomial distribution.
Scheduling and planning
Count possible committees, tournament pairings, tasting flights, or shift rosters before committing to a plan.
Education
Check textbook answers and explore how quickly counts grow as n and r increase.
Planning and decision guide
A quick test for order
Ask whether swapping two chosen items produces a different outcome. “Alice, Bob, Carol” as a committee is the same committee in any order — use a combination. As a podium finish, swapping gold and silver changes the outcome — use a permutation. When in doubt, compute both; their ratio is always r!.
Frequently asked questions
What is the difference between a combination and a permutation?
A combination counts selections where order does not matter; a permutation counts arrangements where order matters. P(n, r) = C(n, r) × r!.
What does C(10, 3) equal?
C(10, 3) = 10! ÷ (3! × 7!) = 120.
Why is C(n, 0) equal to 1?
There is exactly one way to choose nothing, and 0! is defined as 1, so C(n, 0) = n! ÷ (0! × n!) = 1.
Can r be larger than n?
No. Choosing more items than the set contains is impossible without replacement, so the calculator rejects it.
Why is my result an exact integer with no rounding?
Counts are computed with big-integer factorials, so results with hundreds of digits are exact rather than approximated in scientific notation.
Sources and review
- Combination — Wolfram MathWorld. Accessed 2026-08-19.
- Permutation — Wolfram MathWorld. Accessed 2026-08-19.
Reviewed 2026-08-19.