A committee or a podium
Selecting 3 people from 10 gives 120 unordered committees. Assigning those people to three distinct positions gives 720 ordered arrangements without repetition.
Calculations run locally in your browser. This tool does not upload the numbers you enter.
How to use the Combinations and Permutations Calculator
- 01
Enter the number of distinct available items n and the number selected r. Both must be whole numbers from 0 to 1000.
- 02
Check Order matters if changing positions creates a different outcome. Check Repetition allowed if the same item or type may appear again.
- 03
Calculate to see the formula and exact count, then copy or download the full integer.
Examples
Repeated choices
Four positions filled from three symbols give 3^4 = 81 ordered strings. Choosing four items from three types without considering order gives C(6, 4) = 15 multisets.
Order changes the counting question
For combinations, selecting A then B is the same selection as B then A. For permutations, the positions distinguish those outcomes. Without repetition, the formulas are C(n, r) = n!/[r!(n − r)!] and P(n, r) = n!/(n − r)!.
What repetition means
Ordered selection with repetition gives n choices for each of r positions, so its count is n^r. Unordered selection with repetition counts multisets and uses C(n + r − 1, r). This assumes unlimited supply from each type. It is different from arranging a fixed collection of duplicate letters.
Empty selections and exact integers
Selecting zero items gives one empty selection regardless of ordering or repetition settings. A positive selection size from zero available types gives zero when repetition is enabled; without repetition, r cannot exceed n. BigInt products and exact integer division preserve every digit of valid counts.
What to know
- n and r are each capped at 1000. Repetition assumes unlimited availability of each distinct type.
Common questions
Can r be larger than n?
Only when repetition is allowed. Repeated selections can reuse a type, while a selection without repetition cannot contain more items than are available.
Does it list every arrangement?
No. It returns the count. Enumerating a large answer would produce an impractically large list.
Can it arrange letters with fixed duplicate counts?
That is a different problem using factorials divided by repeated-letter factorials. The repetition switch here assumes reusable types rather than a fixed inventory.
Why is the count for r = 0 equal to one?
There is exactly one way to choose nothing: the empty selection. This convention makes the standard counting formulas consistent at their boundaries.