Combination Calculator
Fast combination analysis • 2026 edition
Combination Formula:
Show the calculatorCombination: \(C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\)
Permutation: \(P(n,r) = \frac{n!}{(n-r)!}\)
Factorial: \(n! = n \times (n-1) \times (n-2) \times ... \times 1\)
Combinations count selections where order doesn't matter. For example, selecting 3 books from a collection of 5: C(5,3) = 5!/(3!×2!) = 10 possible groups. Permutations count arrangements where order matters: P(5,3) = 5!/(5-3)! = 60 possible arrangements.
Example: For n=5 and r=3:
- Combinations: C(5,3) = 5!/(3!×2!) = 120/(6×2) = 10
- Permutations: P(5,3) = 5!/(5-3)! = 120/2 = 60
- Factorial: 5! = 5×4×3×2×1 = 120
These formulas are fundamental in probability theory and combinatorics.
Parameters
Advanced Options
Results
| Calculation | Result |
|---|---|
| Combinations C(n,r) | 10 |
| Permutations P(n,r) | 60 |
| Value | Result |
|---|---|
| n | 5 |
| r | 3 |
| n-r | 2 |
| r! | 6 |
Comprehensive Combination Guide
Combinations are selections of objects where the order does not matter. For example, selecting 3 books from a collection of 5 books is a combination because the order in which you select them doesn't matter - you still end up with the same 3 books. The number of combinations of n objects taken r at a time is given by the formula C(n,r) = n!/(r!(n-r)!). Combinations are used in various applications including team formation, lottery draws, and menu selections.
The combination formula calculates the number of ways to select r objects from n distinct objects without regard to order:
Combination: \(C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\)
Permutation: \(P(n,r) = \frac{n!}{(n-r)!}\)
Where:
- \(C(n,r)\) = number of combinations
- \(P(n,r)\) = number of permutations
- \(n\) = total number of objects
- \(r\) = number of objects to select
- \(n!\) = n factorial (n × (n-1) × ... × 1)
The key difference between combinations and permutations is whether order matters:
Factorial (denoted as n!) is the product of all positive integers less than or equal to n:
\(n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1\)
Special cases: 0! = 1 and 1! = 1
- Team Formation: Selecting players for a team
- Lottery Draws: Calculating winning probabilities
- Menu Planning: Choosing items from a menu
- Survey Sampling: Selecting participants
Combination Concepts
A selection of objects where order does not matter.
\(C(n,r) = \frac{n!}{r!(n-r)!}\)
Selections of r objects from n distinct objects.
- Order doesn't matter in combinations
- r cannot exceed n
- C(n,r) = C(n, n-r)
Advanced Concepts
This calculator assumes no repetition allowed.
\(C(n+r-1, r)\)
- Objects can be reused
- Also known as multiset coefficient
- Example: choosing 3 fruits from 4 types
- Combinations ≤ Permutations
- Factorials grow very quickly
- Large values may cause overflow
Combination Learning Quiz
How many ways can you select 3 letters from the set {A, B, C, D} if order doesn't matter?
The answer is A) 4. Since order doesn't matter, we use the combination formula: C(n,r) = n!/(r!(n-r)!), where n=4 (total letters) and r=3 (letters to select). So C(4,3) = 4!/(3!×(4-3)!) = 24/(6×1) = 4. The possible selections are: {ABC}, {ABD}, {ACD}, {BCD}. Notice that ABC is the same as BAC since order doesn't matter.
This problem demonstrates why order doesn't matter in combinations. ABC and BAC are considered the same selection because they contain the same letters. If order mattered, we would use permutations instead, which would give P(4,3) = 4!/(4-3)! = 24 different arrangements.
Combination: A selection where order doesn't matter
Permutation: An arrangement where order matters
Factorial: Product of all positive integers up to n
• Use combination when order doesn't matter
• Use permutation when order matters
• C(n,r) = C(n, n-r)
• Remember: "Combination" sounds like "committee" (selection)
• Think of choosing team members vs arranging people in a line
• Always check if repetition is allowed
• Confusing combination with permutation
• Forgetting that order doesn't matter in combinations
• Not considering whether repetition is allowed
A pizza restaurant offers 8 different toppings. How many different 3-topping pizzas can be ordered if no topping can be repeated?
This is a combination problem because the order in which we select the toppings doesn't matter. A pizza with pepperoni, mushrooms, and olives is the same as a pizza with olives, pepperoni, and mushrooms.
Step 1: Identify the values
Total toppings (n) = 8
Toppings to select (r) = 3
Step 2: Apply the combination formula
C(n,r) = n!/(r!(n-r)!)
C(8,3) = 8!/(3!(8-3)!) = 8!/(3!×5!)
Step 3: Calculate factorials
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320
3! = 3 × 2 × 1 = 6
5! = 5 × 4 × 3 × 2 × 1 = 120
Step 4: Complete the calculation
C(8,3) = 40,320/(6×120) = 40,320/720 = 56
Alternative approach:
We can simplify by canceling common factors:
C(8,3) = (8×7×6)/(3×2×1) = 336/6 = 56
Final Answer: There are 56 different 3-topping pizzas that can be ordered.
This problem illustrates the practical application of combinations in everyday scenarios. The key insight is recognizing that the order of toppings doesn't affect the final pizza. The alternative approach shows how we can simplify the calculation by canceling out common terms in the numerator and denominator.
Selection without replacement: Each topping can only be chosen once
Identical outcomes: Different orders produce the same resultDistinct items: Each topping is unique
• Each topping can only be selected once
• Different orders of same toppings = same pizza
• Use combination for unordered selections
• Look for keywords like "select", "choose", "group"
• Ask yourself: does order change the outcome?
• Simplify calculations by canceling common factors
• Using permutation instead of combination
• Forgetting that order doesn't matter
• Not accounting for the restriction that no topping repeats
FAQ
Q: What's the difference between combination and permutation?
A: The key difference is whether order matters:
Combination: Order doesn't matter. For example, if you're selecting 3 books from a collection of 5 to take on vacation, the order in which you select them doesn't matter - you still end up with the same 3 books. The formula is C(n,r) = n!/(r!(n-r)!)
Permutation: Order matters. For example, if you're arranging 3 books on a shelf from a collection of 5, the sequence matters (Book A first is different from Book A second). The formula is P(n,r) = n!/(n-r)!
Think of it this way: combinations are for selecting (where only the group matters), permutations are for arranging (where sequence counts).
For n=5 and r=3: C(5,3) = 10 while P(5,3) = 60. The permutation count is higher because it considers all possible orderings of each group.
Q: When would I use combinations with repetition allowed?
A: Combinations with repetition allowed use the formula C(n+r-1, r), where you can reuse objects. Here are common scenarios:
1. Candy Selection: Choosing 5 candies from 3 flavors where you can pick multiple of the same flavor. With n=3 flavors and r=5 selections, there are C(3+5-1, 5) = C(7,5) = 21 possible combinations.
2. Stock Portfolio: Allocating investments across different stocks where you can invest in the same stock multiple times.
3. Menu Planning: Ordering 4 items from a menu of 10 dishes where you can order the same dish multiple times.
Without repetition (our calculator's default), the formula is C(n,r) = n!/(r!(n-r)!). With repetition allowed, it's C(n+r-1, r). Note that with repetition, r can exceed n, which isn't possible in the no-repetition scenario.