Permutations and combinations calculator

Compute nPr and nCr from any valid n and r.

What this calculator covers

Use this calculator when you need to count ordered arrangements and unordered groups from the same n-and-r setup.

Keeping nPr and nCr together makes it easier to see when a problem changes from sequencing outcomes to simple group selection.

Frequently asked questions

When should I use permutations versus combinations?
Use permutations when the order of selection matters — such as assigning first, second, and third place in a race. Use combinations when only the membership of the group matters and the order does not — such as choosing three people for a committee.
What do n and r represent?
n is the total number of items in the set you are choosing from, and r is the number of items you are selecting. Both must be non-negative whole numbers, and r cannot exceed n.
Why is nCr always smaller than or equal to nPr?
nCr divides nPr by r!, which collapses all the different orderings of the same selected group into a single count. Since r! is always at least 1, nCr can never exceed nPr, and the two are equal only when r is 0 or 1.
What happens when r equals n?
When every item in the set is selected, nPr equals n! (all arrangements of the full set) and nCr equals 1 (there is only one way to choose all items as an unordered group).

Tool

Run the calculation

Result

RESULT · nPr / nCr

â„–164

Choosing 3 items from 10 gives 720 ordered arrangements and 120 unordered combinations.

Ordered arrangements (nPr)
720
Unordered combinations (nCr)
120

Step-by-step solution

  1. 1.Use permutations when order matters: nPr multiplies the top 3 descending factors from 10 to get 720.
  2. 2.Use combinations when order does not matter: divide the permutation count by r! to collapse duplicate orderings.
  3. 3.That leaves nCr = 120 unique groups.

Walkthrough

Visual walkthrough

Permutations count ordered outcomes. Combinations collapse orderings that represent the same group.

  1. 01

    Count the ordered picks

    10P3 = 720

    For the first slot you have n choices, then n - 1 for the next, and so on until r slots are filled.

  2. 02

    Remove duplicate orderings

    720 / 3!

    Each unordered group appears r! times among the permutations because the same members can be rearranged.

  3. 03

    Read the two counts

    Use nPr for sequenced outcomes and nCr for simple group selection.

    720 permutations · 120 combinations