- Combinations
- 120
- Permutations
- 720
How it works
Selecting everyone can still involve ordering
Choosing all 5 people gives one combination, but placing those same people in 5 distinct positions gives 120 permutations. Selecting the same membership does not eliminate the different arrangements when position still matters.
Deciding whether order matters
Choosing 3 people from 5 for an unordered team gives 5! ÷ (3! × 2!) = 10 combinations. Assigning 3 different positions to people from the same group gives 5 × 4 × 3 = 60 permutations, because changing the order changes the assignment. Both calculations assume distinct people and no repeated selection. Decide whether positions matter before choosing the result; the number of outcomes alone does not establish that every outcome is equally likely.
Permutations & combinations: formula and assumptions
P(n,r) = n!/(n−r)!; C(n,r) = n!/[r!(n−r)!].
Objects are distinct, sampling is without replacement, and r cannot exceed n.
Worked example
Using the Permutations & combinations example:
- Total objects (n)
- 10
- Selected objects (r)
- 3
P(n,r) = n!/(n−r)!; C(n,r) = n!/[r!(n−r)!].
Example result
Combinations: 120
Permutations: 720
These are illustrative inputs. Change the values to explore a different scenario.
Formula references
References explain the underlying methods; they do not imply endorsement or individual professional review of Finflo.
Common questions
When should I use a permutation instead of a combination?
Use permutations when order matters. Use combinations when the same selected objects form one group regardless of their order.
Can the same object be selected more than once?
Not in this model. It assumes distinct objects sampled without replacement, so the selection size cannot exceed the available count.
Prepared by Finflo. See our methodology and editorial policy.