Combination Calculator
Calculate combinations (nCr) or permutations (nPr) from n and r, useful for probability problems and counting arrangements.
Combination Calculator
Calculate combinations (nCr) or permutations (nPr) from n and r, useful for probability problems and counting arrangements.
Enter values and compute the result.
A combination counts how many ways you can choose a group of items from a larger set when the order of selection does not matter. This combination calculator returns nCr (combinations, order ignored) or nPr (permutations, order counts) for any whole numbers n and r, so you can answer “how many different groups are possible?” in one step.
How to use this calculator
- Enter n — the total number of items you are choosing from.
- Enter r — how many items you pick (r cannot be larger than n).
- Choose a mode: nCr for combinations, where the order of the chosen items does not matter, or nPr for permutations, where order does matter.
Worked example
How many ways can you choose 3 people from a group of 10, when the order does not
matter? Use n = 10, r = 3, mode nCr. The “n choose r” formula is:
nCr = n! / ( r! × (n − r)! )
nCr = 10! / (3! × 7!)
= (10 × 9 × 8) / (3 × 2 × 1)
= 720 / 6
= 120
So there are 120 possible groups of 3. If order did matter — say you were
ranking the 3 people first, second, and third — you would switch to nPr and get
10 × 9 × 8 = 720, exactly six times larger because each group of 3 can be
arranged in 3! = 6 orders. For more on counting principles, see the
OpenStax College Algebra section on counting principles.
Frequently asked questions
What is the difference between nCr and nPr?
Use nCr (combinations) when order does not matter, such as picking a committee or a hand of cards. Use nPr (permutations) when order matters, such as filling ranked positions. nPr is always larger because it counts each arrangement of the same group separately.
How do I calculate combinations by hand?
Apply the formula nCr = n! / (r! × (n − r)!). A shortcut is to cancel the larger
factorial first: 10! / 7! collapses to 10 × 9 × 8, then divide by 3! to get
120. Choosing the smaller of r and n − r keeps the arithmetic light.
Why is n choose r the same as n choose (n − r)?
Choosing which r items to include is the same as choosing which n − r items to leave out, so the two counts are always equal — for example, choosing 3 from 10 gives the same 120 as choosing 7 from 10.