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Combinations and permutations

Count selections when order does or does not matter. Objects must be distinct and cannot be selected more than once.

Start with your own numbers

Editable example

An example is filled in. Replace it with your own data to calculate.

Enter the total number of distinct objects and how many to choose. Each object may be selected only once.

Enter number of distinct objects n, for example 5. Use the displayed unit; replace the example with your own value.
Enter number selected r (without replacement), for example 3. Use the displayed unit; replace the example with your own value.

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Example result

Combinations: order does not matter C(n,r)

10ways

Selecting three people uses combinations. Assigning first, second and third place uses permutations because order matters.

Permutations: order matters P(n,r)60 ways

Worked steps

  1. Combinations C(5,3) = 5! ÷ (3! × 2!) = 10
  2. Permutations P(5,3) = 5! ÷ 2! = 60
  3. Permutations count different orders separately; combinations count the same selection only once.

Enter integers from 0 to 100 with r no greater than n. Objects are distinct and cannot be selected twice. Combinations ignore order; permutations count order. Choosing zero objects has one empty selection. Results are exact integers. Repetition, circular arrangements, identical objects and other constraints are unsupported.

Method and scope

Combinations use n! / [r! * (n-r)!], with the entire product r! * (n-r)! in the denominator. Permutations use n! / (n-r)!; a factorial multiplies the positive integers up to its argument, with 0! defined as one.

Selecting zero objects gives one empty selection. Exact integer arithmetic preserves long results rather than approximating them as floating-point numbers.

Enter integers from 0 to 100 with r no greater than n. Repetition, circular arrangements, identical objects and extra restrictions are excluded; a count alone is not a probability.

References:OpenStax: permutations and factorials、OpenStax: combinations and the role of order

What this helps you do

Count selections when order does or does not matter. Objects must be distinct and cannot be selected more than once.

What to enter

Start with number of distinct objects n, then fill the other relevant fields in the units shown. The field hints explain the required format and examples.

How the calculation works

Combinations use n! / [r! * (n-r)!], with the entire product r! * (n-r)! in the denominator. Permutations use n! / (n-r)!; a factorial multiplies the positive integers up to its argument, with 0! defined as one.

Worked example

Choosing three people from five gives ten unordered groups. Assigning three different roles from those five gives 60 ordered selections.

Understanding the answer

Selecting zero objects gives one empty selection. Exact integer arithmetic preserves long results rather than approximating them as floating-point numbers.

Scope and limitations

Enter integers from 0 to 100 with r no greater than n. Repetition, circular arrangements, identical objects and extra restrictions are excluded; a count alone is not a probability.

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