Select two from five
5P2=5×4=20 ordered arrangements, while 5C2=20/2!=10 unordered groups. AB and BA differ as permutations but represent one combination.
Calculate exact permutations (ordered arrangements) and combinations (unordered selections) from n total items and r selected items.
A permutation counts arrangements where order matters. A combination counts selections where order does not matter.
Here n is the total number of distinct items and r is the number selected, without replacement.
The exact product n×(n−1)×…×(n−r+1) counts ordered selections.
nPr = n! / (n−r)!Combinations ignore selection order.
nCr = n! / (r!(n−r)!)Every unordered group of r items can be arranged in r! orders.
nPr = nCr × r!5P2=5×4=20 ordered arrangements, while 5C2=20/2!=10 unordered groups. AB and BA differ as permutations but represent one combination.
10P3=10×9×8=720 and 10C3=720/3!=120.
6C6=1 group containing all items, while 6P6=720 possible orders.
Use permutations for rankings, distinct roles, seating orders, and ordered codes without repetition.
Use combinations for committees, teams, subsets, and unordered selections.
n and r must be whole numbers with n≥0 and 0≤r≤n. There is one way to choose nothing, so nC0=1.
This calculator counts possible arrangements and groups. It does not calculate event probabilities or binomial-distribution probabilities.
An ordered selection of r distinct items from n available items.
An unordered selection of r distinct items from n available items.
n is total items and r is the number selected.
Order matters when different positions, ranks, or roles make AB distinct from BA.
nPr=n!/(n−r)!, the number of ordered selections.
nCr=n!/(r!(n−r)!), the number of unordered selections.
There is exactly one way to choose no items: the empty selection.
There is exactly one group containing every available item.
nPr=nCr×r! because each group can be ordered in r! ways.
No. It returns exact counts only.