Ordered arrangements versus unordered selections
Without replacement, combinations count groups, permutations count ordered arrangements and each group has r! orders. With replacement, ordered sequences use nʳ and unordered selections are multisets; the orders per group vary with repeated-item counts. Large counts are shown as complete integers.
After identifying favorable and total equally likely outcomes, use the Probability Calculator to express their ratio as a chance.
The formula
n is the number of distinct item types and r is the number selected, both whole numbers from 0 to 200. Without replacement each item is used at most once and r ≤ n. With replacement each position can use any type; ignoring order counts multiplicities rather than distinct sequences. Zero selections give one empty outcome in either model; zero types and a positive selection size give no replacement outcomes.
Selecting three people from ten
With n = 10 and r = 3, there are 120 three-person groups. Assigning three different positions creates 720 ordered arrangements. Each group has 3! = 6 orders, and 120 × 6 = 720.
How to use this calculator
- Enter the number of distinct available item types as n and selection size as r.
- Choose whether an item can be selected again. Without replacement r must not exceed n.
- Read the live unordered and ordered counts. Expand the model-specific formulas; repeated groups do not all have r! different orders.
A group and a ranked list are different outcomes
Choosing three committee members ignores the order in which their names were drawn. Choosing a president, secretary, and treasurer assigns different roles, so exchanging two names creates another outcome. These tasks use combinations and permutations respectively.
The items must be distinguishable. Arranging repeated letters in a word needs a repeated-item adjustment and is not the same problem as choosing distinct people.
Choose the replacement rule for the task
Without replacement, a selected item cannot appear again. Choosing three distinct people from ten gives 120 groups and 720 assignments to different roles. This model requires r ≤ n.
With replacement, three positions using ten item types give 10³ = 1,000 ordered sequences. Ignoring order gives C(12, 3) = 220 multisets. A three-digit code allowing repeated digits fits the ordered replacement model, including leading zeros when all ten digits can appear in every position.
Repeated selections change the number of orders
With item types A and B and selection size 3, replacement gives eight ordered sequences and four unordered multisets: AAA, AAB, ABB and BBB. The all-identical groups each have one order, while AAB and ABB each have three. Multiplying every group by 3! would overcount.
The calculator reports that order counts vary in replacement mode instead of presenting r! as a universal multiplier. It counts unconstrained selections of item types; arrangements of a fixed bag with specified repeated-object counts, circular arrangements and required-position rules are separate tasks.
Assumptions & limitations
What this calculation assumes
- Item types are distinct; the selected replacement rule determines whether a type can be selected again.
- Counts are whole numbers between zero and 200.
What to keep in mind
- Fixed repeated-object counts, circular arrangements and additional selection restrictions require different models.
- A count of possibilities alone does not establish that every outcome is equally likely.
Common questions
Which result should I use for a lottery selection?
Use combinations only if the selection ignores order and draws distinct items without replacement. Bonus pools and other rules may require additional calculations.
Why is n choose zero equal to one?
There is one way to select nothing: the empty group.
Are very large results rounded?
No. These counts use exact integer arithmetic and are shown in full within the supported input limit.
Can r be larger than n?
Yes when replacement is allowed. For two item types and three selections, there are eight ordered sequences and four unordered multisets. Without replacement, r cannot exceed n.
Sources & further reading
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