How this calculation works
Count selections from distinct objects without replacement. A permutation counts different orders separately, while a combination treats the same chosen group as one selection. Use combinations for a committee and permutations for ordered positions such as first, second and third.
Worked example
This example uses illustrative inputs. Your current result is shown above.
Selecting 3 from 10 distinct objects gives 10P3 = 10 × 9 × 8 = 720 ordered selections. Divide by 3! = 6 to ignore order: 10C3 = 720/6 = 120 groups.
Assumptions and limits
- n and r are whole numbers with 0 ≤ r ≤ n ≤ 5,000.
- Objects are distinct and cannot be selected more than once.
- These counts do not handle repeated identical objects or replacement; large exact counts are displayed as integer text.
Questions people ask
How do I decide whether order matters?
Ask whether swapping two selected items creates a different outcome. Swapping committee members does not; swapping first and second prize winners does.
Why is choosing zero objects counted as one?
There is exactly one empty selection. Both formulas therefore return one when r is zero, including when n is also zero.