Calculate Unique Combinations: Formula and Examples

This calculator finds the number of unique ways to choose a group from a larger set when order does not matter. Enter the total number of items and how many you are choosing; it returns the combination count, using the standard binomial-coefficient formula.

Formula

The number of combinations of n items taken r at a time is:

C(n,r) = n! / [r!(n − r)!]

Here, n is the total number of distinct available items, and r is the number selected. Both are whole-number counts with no physical unit. The exclamation mark means factorial: for example, 4! = 4 × 3 × 2 × 1 = 24. The calculator’s second output, P(n,r) = n! / (n − r)!, counts ordered arrangements for comparison; it is not the answer when order does not matter.

Worked Example

With the default values, there are 10 items and 3 are chosen. The combination calculation is:

• C(10,3) = 10! / [3!(10 − 3)!]
• C(10,3) = 10! / (3! × 7!)
• C(10,3) = (10 × 9 × 8) / (3 × 2 × 1) = 120

So there are 120 unique combinations. If the three chosen items are arranged in order instead, the count is P(10,3) = 10 × 9 × 8 = 720 ordered arrangements. These are the exact results shown for the calculator defaults.

When to Use Combinations

Use combinations when each selection is defined by which items it contains, not by the order in which they were picked. For example, choosing three people from a group of ten gives 120 possible groups. The group containing Ava, Ben, and Chris is the same group regardless of whether the names are listed Ava–Ben–Chris or Chris–Ava–Ben.

Typical applications include counting possible teams, subsets, committees, and lottery-style selections where a set of chosen numbers is treated as unordered. The formula assumes the available items are distinct and that an item cannot be selected more than once. If repeated selections are allowed, a different counting formula may be required.

Combinations Compared with Permutations

The key question is whether switching the order creates a new outcome. In a combination, it does not; in a permutation, it does. For instance, assigning gold, silver, and bronze places to three people is an ordered problem, while selecting three finalists without ranking them is an unordered one.

• Combination: choose 2 items from A, B, C. The outcomes are AB, AC, and BC, for 3 combinations.
• Permutation: arrange 2 items from A, B, C. The outcomes include AB and BA as different arrangements, giving 6 permutations.

When all selected items are distinct, P(n,r) = C(n,r) × r!. Each combination can be ordered in r! ways. This explains why the permutation count is larger whenever r is greater than 1.

How to Choose the Inputs

Set n to the number of distinct options available before selection, and r to the number selected in each group. For example, if you have 12 different books and want groups of 4, use n = 12 and r = 4. Do not enter the number of groups you hope to make as n; it describes the pool of options.

For the ordinary combination formula, inputs must be whole numbers satisfying 0 ≤ r ≤ n. Selecting zero items has one possible outcome—the empty selection—and selecting all n items also has one outcome. If r is greater than n, it is impossible to select that many distinct items without replacement, so the usual count is zero rather than a valid input for this calculator.

Common Mistakes and Limits

• Counting order by accident: If order is irrelevant, do not treat rearrangements of the same group as separate outcomes.
• Using repeated or indistinguishable items: The basic formula treats every option as distinct and allows each item to be selected at most once. Duplicate-looking objects may need a different model if they are not distinguishable.
• Mixing up n and r: The total pool is n; the size of each chosen group is r.
• Rounding too early: Combination counts are whole numbers. For large inputs, intermediate factorials can be enormous, so use a calculator that evaluates the combination directly rather than rounding factorial values by hand.

The result counts possible selections; it does not say how likely any one selection is. Probability requires a defined set of equally likely outcomes and a count of which outcomes meet the event being studied. Nor does the result account for restrictions such as required members, incompatible choices, or different categories unless the problem is split into appropriate cases.

Frequently Asked Questions

How do you calculate the number of unique combinations?
Use C(n,r) = n! / [r!(n − r)!], where n is the total number of distinct items and r is the number selected. This treats selections with the same members as identical, whatever their order.

How many combinations are there when choosing 2 from 5?
C(5,2) = 5! / (2! × 3!) = 10. There are ten distinct pairs.

Does order matter in a combination?
No. Choosing A and then B is the same combination as choosing B and then A. If order changes the outcome, calculate permutations instead.

Can you choose more items than are available?
Not when each item can be chosen only once. The ordinary combination problem requires r to be no greater than n.

Are combinations the same as subsets?
A selection of r items from n is a subset of size r. If you count every possible subset of every size, including the empty set, the total is 2ⁿ.

Related

• Number Combinations: Formula and Calculator
• How Many Combinations? Calculate n Choose r
• Divisibility Calculator: Check if a Number Divides Evenly

in n | Total items | - | 10 | 0
in r | Items chosen | - | 3 | 0
out combinations | Unique combinations | - | ncr(n,r) | 0
out permutations | Ordered arrangements | - | npr(n,r) | 0