Number Combinations: Formula and Calculator

This calculator works out how many ways you can choose a given number of items from a larger set when order does not matter. Enter the total number of distinct items and how many you want to choose; it returns the number of possible combinations.

Formula

The number of combinations of n distinct items chosen r at a time is C(n,r) = n! / (r!(n − r)!). Here, n is the total number of available items, r is the number selected, and the exclamation mark means factorial: n! is the product of the positive integers from 1 through n, with 0! defined as 1. The result is a count, so it has no unit.

The calculator’s second output is the number of ordered arrangements, or permutations: P(n,r) = n! / (n − r)!. This is useful for comparison, but combinations are the answer when changing the order of the same selected items does not create a new outcome.

Worked Example

With the defaults, there are n = 10 distinct items and r = 3 selections. The combination calculation is:

1. C(10,3) = 10! / (3! × (10 − 3)!)
2. = 10! / (3! × 7!)
3. = (10 × 9 × 8) / (3 × 2 × 1)
4. = 720 / 6 = 120 combinations.

For comparison, the ordered arrangements are P(10,3) = 10 × 9 × 8 = 720 permutations. The calculator therefore reports 120 possible combinations and 720 ordered arrangements for its default inputs.

How Combinations Work

A combination is a selection in which order does not matter. Suppose the items are A, B, C, and D, and you choose two. The pairs AB and BA describe the same selection, so they count once. The possible pairs are AB, AC, AD, BC, BD, and CD: six combinations.

The factorial formula first counts ordered selections, then divides out the repeated orderings. Every group of r distinct items can be arranged in r! ways, so dividing by r! leaves each unordered group counted only once. This is why the combination formula has the extra r! in its denominator compared with the permutation formula.

When to Use Each Calculation

Use combinations when you are selecting a group and the order of its members is irrelevant. Use permutations when positions, sequence, or rank make different orders distinct. For example:

• Choosing three people for a committee from ten: combinations, because the same committee remains the same regardless of selection order.
• Assigning first, second, and third place among ten competitors: permutations, because the order changes the result.
• Choosing three toppings from a list: usually combinations if only the set of toppings matters.
• Creating a three-character code from distinct characters: permutations if each position matters and no character can repeat.

Read the combination result as the number of different groups of the requested size—not the number of individual items, and not the number of ways to put those items in order.

Valid Inputs and Common Mistakes

For the standard formula, n and r must be whole numbers, with 0 ≤ r ≤ n. The calculator requires nonnegative values, but a mathematically meaningful selection also requires the chosen count not to exceed the total. For example, choosing 11 items from a set of 10 is not a valid selection.

• Counting order when it does not matter: This gives a permutation count, which is usually larger. Decide whether rearranging the same selected items makes a genuinely different outcome.
• Forgetting whether repetition is allowed: The displayed formula assumes each item can be selected at most once. Situations allowing repeated choices need a different model.
• Using non-integer values: A count of items must be a whole number. A fractional input such as 2.5 people has no ordinary selection interpretation.
• Rounding too early: Combination counts are integers. For large inputs, use exact integer arithmetic when possible; rounding intermediate factorial calculations can produce an incorrect count.

Repetition and Symmetry

If an item may be selected more than once and order still does not matter, the ordinary combination formula does not apply. For choosing r items from n types with repetition allowed, the count is C(n + r − 1, r). For example, choosing four scoops from three flavors when flavors may repeat is C(3 + 4 − 1, 4) = C(6,4) = 15 possible flavor-count selections. This model counts how many of each type are chosen, not an order of scoops.

A useful check is the symmetry identity C(n,r) = C(n,n − r). Choosing three items to include from ten identifies the same groups as choosing seven items to leave out. The two calculations must give the same answer; if they do not, an input or calculation has gone wrong.

Frequently Asked Questions

How do I calculate combinations of numbers?
Choose the total count n and selection size r, then calculate n! / (r!(n − r)!). Use this when order does not matter and repeated selection is not allowed.

What is the difference between combinations and permutations?
Combinations count selections without regard to order. Permutations count ordered arrangements, so ABC and BAC are different permutations but the same combination.

How many combinations are there when choosing 2 from 10?
C(10,2) = 10! / (2! × 8!) = 45. There are 45 distinct pairs from ten items.

Can I choose more items than are available?
No. The standard selection problem requires 0 ≤ r ≤ n. Choosing more items than the set contains is not possible when each item can be used only once.

Does the formula allow repeated numbers?
No. It assumes the available items are distinct and each can be chosen at most once. If repeated choices are allowed, use a combinations-with-repetition formula or another model suited to the situation.

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