Calculate All Possible Combinations With nCr

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, n, and the number selected, r; it returns the count using the standard combination formula.

Formula

The number of combinations is “n choose r,” written as C(n,r) or as the binomial coefficient n choose r. It is calculated as:

C(n,r) = n! / [r!(n − r)!]

• n is the total number of distinct available items, a nonnegative whole number.
• r is the number of items selected, a whole number from 0 through n.
• ! denotes a factorial: for example, 4! = 4 × 3 × 2 × 1 = 24, and 0! = 1.

The extra calculator output shows permutations, where order matters: P(n,r) = n! / (n − r)!. This helps compare the two counts, but the combinations result is the primary answer.

Worked Example

With the default values, n = 10 total items and r = 3 selected:

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

For comparison, if the order of the three selected items matters, P(10,3) = 10 × 9 × 8 = 720 permutations. The calculator displays both values as whole numbers.

What Counts as a Combination?

A combination is a selection in which changing the order does not create a new result. Imagine choosing three people from a group of ten to form a committee. A committee containing Alex, Blair, and Casey is the same committee regardless of whether you list the names as Alex–Blair–Casey or Casey–Alex–Blair. The calculator counts that group once.

By contrast, assigning those three people to first, second, and third place creates different outcomes when their order changes. That is a permutation. A useful test is to ask whether swapping two selected items changes the outcome. If it does, use permutations; if it does not, use combinations.

For the same n and r, permutations are generally larger because each selected group can be arranged in multiple orders. When r is positive, P(n,r) = C(n,r) × r!. For example, the 120 groups of three chosen from ten can each be ordered in 3! = 6 ways, giving 720 arrangements.

When to Use the Calculator

Use combinations when you want to count possible groups, sets, or selections and the items are distinct. Common mathematical examples include selecting committee members, choosing cards for a hand, picking lottery-number sets for a purely mathematical count, or finding how many subsets of a given size can be made from a collection. The calculation gives a count of possible selections; it does not tell you which selection will occur or how likely a real-world outcome is without an appropriate probability model.

Enter whole numbers for both inputs. The number selected cannot exceed the total number available. For example, selecting four objects from twelve is valid, while selecting thirteen from twelve is not. The formula assumes that each item is distinguishable and that an item cannot be selected more than once.

Reading the Count and Its Limits

The answer is a count, not a measurement, percentage, or probability. For instance, 120 combinations means there are 120 distinct groups of the requested size under the stated assumptions. If you want the probability of a particular event, you need to identify which combinations count as favorable and divide by the total number of equally likely outcomes, if those outcomes are indeed equally likely.

This calculator uses the ordinary combination model: selection is without replacement, and order does not matter. It does not directly handle situations where items can be chosen repeatedly, where some items are identical, or where selections are subject to extra rules. Those situations require a different counting method. For choosing r items from n types when repeats are allowed and order still does not matter, a common formula is C(n + r − 1, r), provided that this model matches the problem.

Large inputs can produce enormous counts. Factorials grow quickly, so calculating each factorial separately by hand may be inefficient and can lead to arithmetic errors. The calculator evaluates the combination directly; for a manual calculation, cancel common factors or use a calculator with integer or scientific-notation support. A displayed whole-number result is exact for the formula, though very large results may be difficult to interpret without separators or scientific notation.

Common Mistakes

• Counting order when it should not matter: A group of three people remains the same group in every listing order, so use combinations for group membership.
• Ignoring order when it does matter: Ranked places, ordered schedules, and ordered passwords can distinguish arrangements; combinations alone undercount those outcomes.
• Entering impossible values: Use nonnegative whole numbers and ensure 0 ≤ r ≤ n. Fractional selections have no ordinary meaning in this model.
• Assuming repeats are allowed: The standard formula selects distinct items without replacement. A rule allowing repeated picks changes the problem.
• Confusing a count with a probability: The combination total describes how many groups exist, not the chance of any one group unless the sampling assumptions are specified.

Frequently Asked Questions

How do I calculate all possible combinations?
Identify the total number of distinct items, n, and the number selected, r. Calculate n! divided by r!(n − r)!, or enter those values in the calculator above.

What is the difference between combinations and permutations?
Combinations treat selections with the same members as identical, whatever their order. Permutations count different orders as different outcomes.

How many combinations are there when selecting zero items?
There is exactly one: the empty selection. This follows from C(n,0) = 1, because there is one way to choose nothing from a set.

Can the number selected be larger than the total number of items?
No, not in the standard without-replacement model. The input must satisfy r ≤ n; otherwise there are no valid selections under the stated rules.

Does this formula allow choosing the same item more than once?
No. It counts selections without replacement. If repeats are allowed, use a formula designed for that situation and confirm whether order matters.

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in n | Total items (n) | - | 10 | 0
in r | Items selected (r) | - | 3 | 0
out combinations | Possible combinations | - | ncr(n,r) | 0
out permutations | Arrangements if order matters | - | npr(n,r) | 0