How Many Combinations? Calculate n Choose r

This combinations calculator finds how many different groups of a chosen size can be selected from a larger set when order does not matter. Enter the total number of items and the number selected to get the exact count of possible combinations.

Formula

For choosing r items from n distinct items without regard to order, use:

C(n,r) = n! / [r!(n-r)!]

• n is the total number of distinct available items, a count.
• r is the number of items selected, a count.
• The exclamation mark means factorial: n! is the product of all positive integers from 1 through n, with 0! defined as 1.

The calculator also shows the number of ordered arrangements as a comparison: P(n,r) = n!/(n-r)!. Both results are counts, so they have no physical units. The combination formula is also written “n choose r” and as the binomial coefficient ⁿCᵣ.

Worked Example

With the default values, there are 10 total items and 3 are selected. Since order does not matter:

1. Substitute into the formula: C(10,3) = 10! / [3!(10-3)!].
2. Cancel the shared 7! factor: (10 × 9 × 8) / (3 × 2 × 1).
3. Calculate: 720 / 6 = 120 combinations.

The comparison output counts ordered arrangements instead: P(10,3) = 10 × 9 × 8 = 720 arrangements. The calculator displays both results as whole numbers, matching the formula’s exact values.

What Counts as a Combination?

A combination is a selection in which the membership of the group matters, but the order of its members does not. Imagine selecting three people from a group of ten to form a committee. The group containing Ana, Bo, and Cal is the same committee whether you list the names as Ana–Bo–Cal or Cal–Ana–Bo.

Those six possible orderings of the same three people are not six different committees. The permutation count treats them separately; the combination count treats them as one. That is why, when 1 ≤ r ≤ n, the number of combinations is the number of permutations divided by r!.

• Choose a team, committee, or set of ingredients: typically a combination.
• Assign first, second, and third places: an ordered arrangement, so use permutations.
• Choose a lock code or sequence: order usually matters, though repetition rules must also be specified.

How to Use the Result

Enter whole numbers for both inputs. The total, n, is the size of the original set; r is the size of each group being counted. For the standard formula, use distinct items and select without replacement: an item cannot appear twice in one group.

For example, choosing 3 books from a shelf of 10 gives 120 possible sets of books. If the goal instead is to place 3 different books in a particular left-to-right order, there are 720 arrangements. Decide what makes two outcomes different before relying on either number.

The combinations result is the direct answer when order is irrelevant. The permutations output is included only to make the order distinction clear; it is not an additional part of the combination count.

Valid Inputs and Common Mistakes

For ordinary selections, the inputs must be whole numbers satisfying 0 ≤ r ≤ n. A selection of zero items has one possibility: the empty selection. Selecting every item also has one possibility. If r is greater than n, the requested group cannot be formed, so the standard combination formula does not apply.

• Counting order accidentally: If rearranging the selected items does not make a new group, use combinations rather than permutations.
• Using non-integer values: Counts of items must be whole numbers. A value such as 2.5 items has no meaning in this model.
• Rounding too early: Large factorials grow quickly. Use the combination formula or a calculator designed for it instead of rounding intermediate factorial values.
• Assuming items are distinct when they are not: The formula treats each available item as individually distinguishable. Repeated types or identical objects can require a different counting method.
• Ignoring replacement: The formula assumes an item cannot be selected more than once in a group. Selection with replacement uses a different formula.

Limits and Related Counting Rules

The calculator covers selections of distinct items without replacement. If order matters, use permutations. If order does not matter but repeats are allowed, the count for selecting r items from n types is C(n+r-1,r), provided there is at least one type and the usual indistinguishable-selection model applies.

Some problems also impose restrictions, such as requiring a particular item, forbidding two items from appearing together, or dividing selections into categories. The basic formula does not account for those conditions by itself. Such problems may be handled by splitting the cases into valid groups, using complementary counting, or applying another combinatorial rule.

For a useful check, the number of ways to choose r items equals the number of ways to leave out n-r items: C(n,r) = C(n,n-r). Choosing three committee members from ten therefore gives the same count as choosing the seven people who are not on the committee.

Frequently Asked Questions

How many combinations are there when choosing 3 from 10?
There are 120 combinations, calculated as 10! divided by 3! times 7!.

What is the difference between a combination and a permutation?
A combination ignores the order of selected items. A permutation counts different orders as different outcomes.

Can the number of combinations be bigger than the number of items?
Yes. A set of 10 items has 120 different groups of 3, even though it contains only 10 individual items.

Why does choosing 2 from 5 equal choosing 3 from 5?
Each pair left out uniquely identifies the three items chosen, and each group of three uniquely identifies the two left out. Thus C(5,2) = C(5,3) = 10.

Does this formula allow selecting the same item more than once?
No. It counts selections without replacement. If repeats are allowed, the counting rule depends on whether order matters and whether repeated items are treated as distinct.

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