A combinations with repetition calculation answers a simple counting question: how many groups can you form if you may reuse items and the order of selection does not matter? This is common when choosing flavors, toppings, symbols, or any set of items where repeated picks are allowed.

The standard formula is based on ordinary combinations. If you have n item types and choose r items with repetition allowed, the count is \( \binom{n+r-1}{r} \), often read as “n plus r minus 1 choose r.” You may also see this written using factorials or in terms of the ordinary combination calculator as a related counting method.

What combinations with repetition means

A combination is a selection where order does not matter. With repetition, you are allowed to use the same item more than once. That means these selections are considered the same:

  • A, A, B
  • A, B, A
  • B, A, A

All of them represent the same combination when order does not matter.

This is different from permutations, where order does matter. If you need to compare the two ideas, a combination and permutation calculator can help you see the difference quickly.

When people use this calculation

You use combinations with repetition when you are selecting a fixed number of items from a set, and both of these are true:

  1. The order of the selection does not matter.
  2. You can choose the same type more than once.

Typical examples include:

  • choosing scoops of ice cream from a list of flavors
  • choosing repeated ingredients or toppings
  • forming multisets in counting problems
  • selecting symbols, categories, or resource types with reuse allowed

If you are dealing with a random process rather than a pure counting problem, that is a different topic. For related math tools, you may also find the number theory calculator useful for other discrete math tasks.

The formula

The number of combinations with repetition from n types taken r at a time is:

\[ \binom{n+r-1}{r} \]

You may also see this written as:

\[ \binom{n+r-1}{n-1} \]

Both forms are equivalent.

Why the formula works

Think of the problem as distributing r identical selections into n categories. A standard counting argument turns that into an ordinary combination problem by adding separators. The result is the same count as choosing r positions from a total of n+r-1 positions.

In factorial form, the same calculation is:

\[ \binom{n+r-1}{r} = \frac{(n+r-1)!}{r!(n-1)!} \]

That is the direct method behind most calculators, including this one. If you want a broader overview of calculator behavior and input/output conventions, see the Wikipedia article on calculator devices.

How to read the result

The output is the number of distinct unordered selections you can make under the rules above. Each distinct grouping counts once, even if it can be formed in several different orders.

For example, if the result is 10, that means there are 10 different multisets or repeated-choice combinations. It does not mean there are 10 ways to arrange the items, and it does not mean 10 total objects were chosen.

A useful way to interpret the result is:

  • small result: only a few unique repeated selections are possible
  • larger result: many distinct groups can be formed from the same item types

If you are comparing several counting methods, a random number generator is not relevant here because this is exact counting, not simulation.

Worked examples

Example 1: 3 flavors, choose 2 scoops

Suppose you have n = 3 flavors and you choose r = 2 scoops. The count is:

\[ \binom{3+2-1}{2} = \binom{4}{2} = 6 \]

So there are 6 distinct combinations with repetition.

You can list them as:

  • AA
  • AB
  • AC
  • BB
  • BC
  • CC

This matches the formula result.

Example 2: 4 item types, choose 3 items

Suppose n = 4 and r = 3. Then:

\[ \binom{4+3-1}{3} = \binom{6}{3} = 20 \]

So the number of combinations with repetition is 20.

You can think of this as all unordered triples from four types, where repeats are allowed. Examples would include:

  • AAA
  • AAB
  • ABB
  • ABC
  • CCC

Each unique pattern counts once.

Example 3: 5 categories, choose 4 items

Suppose n = 5 and r = 4. Then:

\[ \binom{5+4-1}{4} = \binom{8}{4} = 70 \]

So the result is 70 distinct selections.

This kind of problem often appears when you are counting repeated picks from a menu of categories, such as four identical choices drawn from five possible types.

Practical interpretation and common uses

The formula is not limited to “objects in a bag” examples. It applies whenever repeated choice is allowed and the only thing that matters is how many of each type you picked.

Here are common real uses:

  • counting ways to choose a set of toppings when repeats are allowed
  • counting combinations of assets, categories, or labels in a model
  • counting nondecreasing selections in a math problem
  • determining the number of possible multisets of size r from n types

If you are working with totals, averages, or grouped quantities after the count, other calculators may help with the next step, such as an average calculator.

Common mistakes

Treating order as important

The biggest mistake is counting different orders separately. For combinations with repetition, AAB, ABA, and BAA are the same selection. If order matters, you need a different method.

Forgetting that repetition is allowed

Some people use ordinary combinations by habit and forget that repeated choices change the count. If repeats are allowed, the count is usually larger than ordinary combinations.

Mixing up n and r

  • n = number of item types available
  • r = number of items you choose

If you swap them, the formula gives the wrong value.

Using the wrong formula for impossible cases

This calculator is for counting selections with nonnegative integer inputs. If either value does not fit the problem setup, the result is not meaningful. For example, if you are counting a process with ordering, constraints, or dependence between picks, this formula is not the right tool.

Confusing “with replacement” and “with repetition”

These phrases are often used in the same counting context, but the meaning should stay clear: you can use the same type again after a choice. The key question is whether the same item type can be picked more than once.

Limits and what the calculator does not do

This calculation gives an exact count for a specific combinatorics setup. It does not:

  • list every combination
  • tell you probabilities
  • account for weights or unequal chances
  • handle order-sensitive arrangements
  • handle extra restrictions like “no more than two of any type”

If your problem includes restrictions such as mandatory items, limited repeats, or separate categories, you may need a more specialized counting method. For simple exact selection counts, the formula above is the standard one.

How to use the result in your own problem

Once you know the number of types n and the number chosen r, plug them into:

\[ \binom{n+r-1}{r} \]

Then interpret the answer as the number of unique unordered selections with repetition allowed. If the result seems too high or too low, recheck whether your problem really allows repeats and whether order is irrelevant.

If you want to compare this with other counting ideas, try checking the same setup with a permutation-style interpretation or a standard combination calculation to make sure the model matches the problem.

Frequently asked questions

What does “with repetition” mean in combinations?
It means you can choose the same type more than once. The count is based on unique groups, not on the order in which you picked them.
Why is the formula \(\binom{n+r-1}{r}\)?
The extra \(r-1\) comes from converting a repeated-choice problem into a standard combination problem using separators. That is why the count is larger than ordinary combinations.
Is AAB the same as ABA?
Yes, if order does not matter. In combinations with repetition, both represent the same selection because they contain the same item counts.
When should I not use this formula?
Do not use it when order matters, when repeats are not allowed, or when the problem has extra restrictions that change the counting. In those cases, you need a different counting method.
Can the result ever be zero?
Not in the usual valid counting setup with nonnegative integers and a defined choice problem. If the inputs do not match a real selection scenario, the formula may not apply.
What is the difference between ordinary combinations and combinations with repetition?
Ordinary combinations count selections without repeats, while combinations with repetition allow the same type to be chosen more than once. That makes the repeated-choice count larger in most cases.
Can I use this for choosing identical items from categories?
Yes. If you are choosing a fixed number of items from named categories and order does not matter, this is the right counting method. Check that repeats are allowed before you use it.

About this article

Written by Yahav Atia, WSO – the digital agency (wso.co.il) that owns and founded this site, with the help of AI, and passed automated quality checks before publication. Numbers in the worked examples were computed in code, not estimated.