Combinations Without Repetition: Formula and Calculator

This calculator finds the number of ways to choose a specified number of distinct items from a larger group when you cannot choose an item more than once and the order does not matter. Enter the total number of items, n, and the number chosen, r; the result is the combination count, written as n choose r.

Formula

The number of combinations without repetition is the number of ways to choose r items from n distinct items when order is ignored:

C(n, r) = n! / (r! × (n − r)!)

• n is the total number of available distinct items, a nonnegative whole number.
• r is the number of items selected, a whole number from 0 through n.
• ! means factorial: n! is the product of the positive integers from 1 to n, with 0! defined as 1.

The answer is a count, so it has no unit. The calculator also shows nPr as a comparison: nPr(n, r) = n! / (n − r)!, the count when the order of the selected items does matter.

Worked Example

With the calculator’s default values, there are n = 10 items and r = 3 are selected. Because order does not matter, choosing A, B, C is the same group as choosing C, B, A.

1. Start with C(10, 3) = 10! / (3! × 7!).
2. Cancel 7! from the numerator and denominator: (10 × 9 × 8) / (3 × 2 × 1).
3. Calculate 720 / 6 = 120 combinations.

For comparison, the ordered count is 10P3 = 10 × 9 × 8 = 720 permutations. The combination result is the calculator’s headline answer; 720 is a separate comparison value.

What “Without Repetition” Means

Without repetition means that each item can appear at most once in a selection. If a group has five different books, a selection of three books cannot include the same book twice. “Order does not matter” is a separate condition: the set of books on the shelf is the same regardless of the order in which they were picked.

These two conditions distinguish combinations without repetition from other counting problems:

• Combination without repetition: choose items once each; ignore order.
• Permutation without repetition: choose items once each; count different orders as different outcomes.
• Combination with repetition: ignore order, but an item or category may be chosen more than once.

For example, choosing three people for a committee is ordinarily a combination. Assigning those three people to president, secretary, and treasurer is a permutation, because swapping two roles creates a different assignment.

How to Choose the Right Calculation

Before using the formula, identify what counts as an item and what makes two outcomes different. Use combinations when a result is just a group, hand, or set, with no meaningful positions or sequence. Use permutations when order, rank, or assigned roles changes the outcome.

• Choosing 3 toppings from 10 distinct options, with no repeats: C(10, 3) = 120.
• Choosing first, second, and third place from 10 competitors: 10P3 = 720.
• Choosing 3 scoops from 10 flavors when a flavor can be repeated and scoop order is irrelevant: this is a combination with repetition, not the formula above.

A useful check is to ask: if I swap the selected items’ order, is it still the same outcome? If yes, combinations may fit. Then ask whether any item can be selected more than once. If no, use combinations without repetition.

Useful Properties and Common Mistakes

Several checks help catch errors. First, r must be a whole number between 0 and n, inclusive. A selection of more distinct items than are available is impossible, and fractional item counts do not describe an ordinary selection problem.

• Choosing none: C(n, 0) = 1. There is one way to choose an empty group.
• Choosing everything: C(n, n) = 1.
• Symmetry: C(n, r) = C(n, n − r). Choosing which r items to take is equivalent to choosing which n − r items to leave out.
• Order factor: nP r = C(n, r) × r!, because each chosen group can be arranged in r! orders.

A common mistake is to use nPr when the question asks only for groups. That counts every ordering repeatedly. Another is to treat identical-looking objects as distinct without checking the problem: the standard formula assumes n distinguishable items. If some objects are identical, extra care is needed because exchanging identical objects may not create a new outcome.

For large n, factorials become very large, so calculating n! and the denominator separately can cause overflow in some software or calculators. A reliable combination function or a cancellation-based calculation is preferable. Also, do not round intermediate values in a more complicated counting problem; rounding can change an exact count. The final answer should be a whole number.

Frequently Asked Questions

What is the formula for combinations without repetition?
C(n, r) = n! / (r! × (n − r)!). It counts selections of r distinct items from n when order does not matter.

How do I calculate n choose r?
Enter n as the total number of available items and r as the number selected. Apply the formula, making sure both are whole numbers and 0 ≤ r ≤ n.

Does order matter in a combination?
No. A combination counts a selected group once, regardless of the order in which its items are listed or chosen.

What is the difference between a combination and a permutation?
A combination counts groups; a permutation counts ordered selections. For the same n and r, the permutation count is the combination count multiplied by r!.

Can r be zero or equal to n?
Yes. There is exactly one way to choose no items and exactly one way to choose all n items, so C(n, 0) = C(n, n) = 1.

What if an item can be selected more than once?
Then this formula does not apply as stated. If order is still irrelevant and repetition is allowed, a different formula is used: C(n + r − 1, r), assuming n available item types and r selections.

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in n | Total items (n) | - | 10 | 0
in r | Items chosen (r) | - | 3 | 0
out combinations | Combinations without repetition | - | ncr(n,r) | 0
out permutations | Ordered selections for comparison | - | npr(n,r) | 0