Choosing 2 from 6
From 6 items, there are 30 permutations of 2 (order matters) but only 15 combinations (order does not).
Permutation and combination calculator
Result
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How it works
Both formulas start from n! (n factorial), the number of ways to arrange all n items. Permutations divide out only the arrangements of the items not chosen, leaving every ordering of the r chosen items as distinct. Combinations go one step further and also divide out the r! ways to order the chosen group itself, since order does not matter — which is exactly why combinations are always the smaller of the two numbers.
The ratio between them is always r! exactly: every combination corresponds to r! different permutations, one for each way of ordering that same group.
Formulas used
Permutations
nPr = n! ÷ (n − r)!
Combinations
nCr = n! ÷ (r! × (n − r)!)
Worked examples
A larger set
Choosing 3 from 10 gives 720 permutations and 120 combinations — a ratio of exactly 6, since 3! = 6.
Assumptions and limits
- Items are treated as distinct — no two items are identical.
- Each item can be selected at most once (no repetition).
Frequently asked questions
How do I know whether to use permutations or combinations?
Ask whether order matters for the problem: assigning 1st, 2nd and 3rd place in a race is a permutation (order matters); picking a 3-person committee from a group is a combination (order does not matter, only who is on it).
Why is nCr always smaller than nPr (when r > 1)?
Because every combination gets counted r! times over in the permutation count — once for each way of ordering the same r items — so dividing by r! collapses those duplicates into a single combination.
What if r equals n?
nPr becomes n! (every full arrangement of all the items), and nCr becomes exactly 1 (there is only one way to "choose" all of them, since order does not matter).
What if r is 0?
Both nPr and nCr equal 1 — there is exactly one way to choose nothing at all.
Updated
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