Combinations with Replacement Calculator
Find CR(n,r) instantly. Enter the population size n and sample size r to count multiset selections where repetition is allowed and order does not matter.
| Step | Value |
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CR(n,r) Quick Reference Table
The combinations with replacement reference table below lists common CR(n,r) values for n from 1 to 6 and r from 0 to 5. Use it to cross-check your result or spot patterns.
| n \ r | r = 0 | r = 1 | r = 2 | r = 3 | r = 4 | r = 5 |
|---|---|---|---|---|---|---|
| n = 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| n = 2 | 1 | 2 | 3 | 4 | 5 | 6 |
| n = 3 | 1 | 3 | 6 | 10 | 15 | 21 |
| n = 4 | 1 | 4 | 10 | 20 | 35 | 56 |
| n = 5 | 1 | 5 | 15 | 35 | 70 | 126 |
| n = 6 | 1 | 6 | 21 | 56 | 126 | 252 |
What Is Combinations with Replacement?
Combinations with replacement counts the number of ways to select r items from a set of n distinct types when the same type can appear more than once and order does not matter. It is also called a multiset coefficient or multichoose.
The key difference from standard combinations is that you are drawing from the full set each time. If you choose an apple from a basket of fruit, it goes back before the next draw, so you could choose apple again.
Choose r from n, no repeats, order does not matter. Each item can only appear once in a selection.
Choose r from n, repeats allowed, order does not matter. An item can be selected multiple times in one set.
Choose r from n, no repeats, order matters. Different arrangements of the same items count separately.
Choose r from n, repeats allowed, order matters. Each slot independently picks from all n types.
How to Use This Calculator
This combinations with replacement calculator requires just two inputs to produce a complete result with a step-by-step factorial breakdown.
- Enter n, the total number of distinct object types in your set.
- Enter r, the number of items you want to select (r can exceed n).
- Press Calculate CR(n, r) or hit Enter to compute.
- Read the result and review each factorial step in the table below.
- Press Clear to reset and try a new calculation.
Worked Example
A pizza shop offers 5 toppings and you choose 3 toppings where repeats are allowed. Order does not matter, so extra cheese counted once is the same as extra cheese counted last.
CR(5,3) = (5 + 3 − 1)! ÷ (3! × 4!) = 7! ÷ (6 × 24) = 5040 ÷ 144 = 35
There are 35 distinct topping combinations when repeats are allowed.
Combinations with Replacement Formula
The combinations with replacement formula is derived from the standard binomial coefficient by substituting n + r − 1 for the upper parameter.
Why This Formula Works
Imagine placing r identical balls into n labeled boxes. Every such arrangement maps one-to-one onto a multiset of size r from n types. The number of ways to do this is equivalent to choosing r positions from n + r − 1 total slots, which is the standard formula above.
Special Cases
- When r = 0, CR(n,0) = 1 for any n. There is exactly one empty selection.
- When n = 1, CR(1,r) = 1 for any r. Only one type exists so all r items are identical.
- When n = 0 and r = 0, CR(0,0) = 1 by convention.
- When n = 0 and r is greater than 0, CR(0,r) = 0. No objects means no selection.
Relationship to Pascal Triangle
Because CR(n,r) = C(n+r-1, r), every value in the combinations with replacement table appears somewhere in Pascal’s Triangle at row n+r-1 and column r. This connection provides a fast manual check for small values.
Real-World Uses of CR(n,r)
Combinations with replacement appears across many fields wherever repeated selection from a fixed set of options must be counted without regard to order.
Food and Consumer Choices
Choosing r scoops of ice cream from n flavors, selecting r toppings from n menu items, or filling r slots in a gift box from n product types. All involve multiset counting.
Probability and Statistics
Sampling with replacement from a finite population, computing multinomial coefficients, or setting up Bayesian update tables where category counts can repeat across draws.
Computer Science
Counting the number of non-decreasing sequences of length r over an alphabet of size n, allocating identical tokens across categories, and analyzing hash collision probabilities in data structures.
Chemistry and Physics
Distributing r indistinguishable particles across n energy states (Bose-Einstein statistics), counting the number of monomers in polymer chains, and enumerating molecular isomers.
Game Design
Computing hand sizes from a deck where cards can repeat, counting possible item loadouts when duplicates are allowed, or enumerating tile combinations in board games.
Frequently Asked Questions
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Combinations with replacement is a counting method that finds the number of ways to choose r items from a set of n distinct objects when order does not matter and each object can be chosen more than once. It is also called a multiset or multichoose.
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The formula for combinations with replacement is CR(n,r) = (n + r − 1)! divided by (r! times (n − 1)!). This is equivalent to C(n+r-1, r), the standard binomial coefficient applied to n+r-1 and r.
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C(n,r) counts combinations without replacement, meaning each item can only be chosen once. CR(n,r) counts combinations with replacement, meaning items can repeat. CR(n,r) always produces a result greater than or equal to C(n,r) for the same n and r values.
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In the combinations with replacement formula, n is the total number of distinct objects or types to choose from (the population size), and r is the number of items you select (the sample size or subset size). Both n and r must be non-negative integers.
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Yes. Because replacements are allowed, r can be larger than n. For example, if n = 3 flavors of ice cream and r = 5 scoops, you can still compute CR(3,5) = 21 valid combinations even though you are choosing more scoops than there are flavors.
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Suppose a candy shop has 4 types of sweets (n = 4) and you pick 3 pieces (r = 3), allowing duplicates. The number of distinct selections is CR(4,3) = (4+3-1)! divided by (3! times 3!) = 6! divided by (6 times 6) = 720 divided by 36 = 20 different combinations.
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Permutations with replacement count ordered arrangements where repetition is allowed, giving n^r outcomes. Combinations with replacement count unordered selections where repetition is allowed, giving CR(n,r) outcomes. Combinations always produce fewer results because different orderings of the same items are treated as one selection.
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When r = 0, there is exactly 1 combination (the empty selection), so CR(n,0) = 1 for any n. When n = 0 and r = 0, CR(0,0) = 1 by convention. When n = 0 and r is greater than 0, CR(0,r) = 0 because there are no objects to choose from.