Permutation & Combination Calculator
Permutations and combinations both count ways to select objects from a set. The deciding question is whether order matters.
Use n for the total number of distinct objects and r for the number selected. A combination ignores the order of the selected objects; a permutation treats different orders as different outcomes.
01 · The key ideaOrder Matters or Not?
Order is ignored
Order creates a new result
02 · DefinitionsFour Common Counting Cases
What changes the formula?
| Case | Order | Replacement | Typical notation |
|---|---|---|---|
| Combination | Does not matter | No | nCr |
| Permutation | Matters | No | nPr |
| Combination with replacement | Does not matter | Yes | C(n+r−1,r) |
| Permutation with replacement | Matters | Yes | nr |
03 · FormulasPermutation and Combination Formulas
The combination formula removes duplicate orderings from the permutation count. The relationship can be written as:
Notation at a glance
| Symbol | Meaning | Example |
|---|---|---|
| n | Total number of distinct objects | 10 people |
| r | Number selected or arranged | 3 people |
| n! | Factorial of n | 5! = 120 |
| nPr | Ordered selections | 10P3 = 720 |
| nCr | Unordered selections | 10C3 = 120 |
04 · Worked exampleChoosing 3 from 10
Imagine 10 distinct people. Selecting 3 people for a committee ignores their order; assigning those 3 people to three different positions does not.
10C3 = 10! / (3! × 7!) = 120
Permutation:
10P3 = 10! / 7! = 720
The permutation count is six times larger because the same three selected people can be arranged in 3! = 6 different orders.
05 · RepetitionWhen Objects Can Be Used More Than Once
Allowing repetition changes the number of available choices at each position. For ordered selections, every position has n choices.
Example with n = 5 and r = 3
| Type | Calculation | Result |
|---|---|---|
| Permutation, no repetition | 5P3 | 60 |
| Combination, no repetition | 5C3 | 10 |
| Permutation, repetition | 5³ | 125 |
| Combination, repetition | 5H3 = C(7,3) | 35 |
06 · Classic applicationHandshake Problem
If every pair in a group of n people shakes hands exactly once, the order of the two people does not matter. That makes this a combination problem: choose 2 people from n.
= 10 × 9 / 2
= 45 handshakes
07 · More examplesWhich Formula Fits?
Common counting problems
| Problem | Use | Answer |
|---|---|---|
| Choose 2 prizes from 6 | Combination | 6C2 = 15 |
| Choose 3 students from 25 | Combination | 25C3 = 2,300 |
| Assign 2 distinct roles from 11 people | Permutation | 11P2 = 110 |
| Choose 4 favorites from 18 menu items | Combination | 18C4 = 3,060 |
| Three positions, 10 choices each, repeats allowed | Permutation with repetition | 10³ = 1,000 |
08 · Pascal's triangleHow nCr Values Form a Pattern
Combination values appear in Pascal's triangle. For a fixed n, the row lists C(n,0), C(n,1), C(n,2), and so on through C(n,n).
Selected rows
| n | Values of C(n,r) |
|---|---|
| 0 | 1 |
| 1 | 1, 1 |
| 2 | 1, 2, 1 |
| 3 | 1, 3, 3, 1 |
| 4 | 1, 4, 6, 4, 1 |
| 5 | 1, 5, 10, 10, 5, 1 |
| 10 | 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1 |
09 · Practical checkConditions and Edge Cases
Standard without-replacement formulas
| Condition | Result |
|---|---|
| r = 0 | nCr = 1 and nPr = 1 |
| r = n | nCr = 1 and nPr = n! |
| 0 < r < n | Both counts are valid; nPr ≥ nCr |
| r > n | No valid selection without repetition |
10 · FAQFrequently Asked Questions
Think of a permutation as a lineup: changing the order changes the result. Think of a combination as a group: changing the order does not.
For the same selected objects, a permutation counts their possible orders. A combination treats those orders as duplicates.
For the standard without-replacement formulas, no selection is possible when r > n. Repetition uses different formulas.
Yes. Both are standard counting tools for probability problems where the number of possible outcomes must be determined.