LifeTally
Selections without repetition
How many ways can you choose k of n distinct objects, with and without order?
Compare the exact count of ordered lists and unordered selections of the same size.
Your distinct objects and selection size
Every object is distinct and may appear at most once. Choose exactly k objects. Swapping two positions creates a different ordered list, but leaves the unordered selection unchanged.
How the estimate works
The first position has n choices, the next n − 1, continuing for k positions. Thus P(n,k) = n! ÷ (n − k)!. Every k-object selection has k! orders, so C(n,k) = n! ÷ [k! × (n − k)!]. Both valid counts are required for both results, including k = 0. Here m! means 1 × 2 × … × m for a positive whole m, and 0! = 1.
P(n,k) = n! ÷ (n − k)!
C(n,k) = n! ÷ [k! × (n − k)!]
0! = 1; P(n,0) = C(n,0) = 1The limit n ≤ 1000 is our processing and display limit, not a mathematical limit or recommended size. This model excludes repetition, replacement, indistinguishable objects, circular arrangements, partitions, forbidden pairs and other selection restrictions. It counts outcomes without listing or randomly drawing them. It does not calculate probabilities, lottery odds or password security; counts do not establish equally likely outcomes or real-world feasibility.
OpenStax: permutations without repetition ↗OpenStax: combinations and order ↗
Check the filled example
Our chosen example: n = 11 and k = 4. There are 11 × 10 × 9 × 8 = 7920 ordered lists. Each four-object selection has 4 × 3 × 2 × 1 = 24 orders, so 7920 ÷ 24 = 330 unordered selections. These are example counts, not a recommendation.
Accepted input values
- Available distinct objects n
- 11
- Objects selected k
- 4
Plan the next step
Decide whether positions distinguish outcomes in your question, then use the corresponding result. Use the standard calculator for separate arithmetic; a count alone is not a probability.
Open standard calculator →