Combinations Calculator
Calculate combinations (nCr) with step-by-step solutions, Pascal's Triangle visualization, and practical examples like lottery odds and team selection.
Results
Combinations
120
C(10, 3)
Input Values
Total number of items to choose from
Number of items to select
CR(n,r) = C(n+r-1, r)
Visual representation (limited to n <= 12)
Results
Standard Combinations
Ways to select 3 items from 10 (order does not matter)
120
C(10, 3)
Combinations vs Permutations
Combinations (order does not matter)
120
Permutations (order matters)
720
Ratio: P(10,3) / C(10,3) = 6 = 3!
Step-by-Step Calculation
Pascal's Triangle
C(10, 3) = 120 is highlighted at row 10, position 3
Property: Sum of row 10 = 1,024 = 210
Practical Examples
Click to calculate:
Lottery Odds Calculator
Your Current Selection
Picking 3 numbers from 10
1 in 120
Probability: 0.83333333%
Common Lotteries
Combination Formulas
Standard Combination
C(n, r) = n! / (r! x (n-r)!)
Order does not matter, no replacement
With Replacement
CR(n, r) = C(n+r-1, r)
Order does not matter, with replacement
Symmetry Property
C(n, r) = C(n, n-r)
C(10, 3) = C(10, 7)
Pascal's Identity
C(n, r) = C(n-1, r-1) + C(n-1, r)
Each Pascal entry is sum of two above
Results
Combinations
120
C(10, 3)
?How Do You Calculate Combinations?
Combinations count selections where order does NOT matter. Formula: C(n,r) = n! / (r! x (n-r)!). For example, choosing 3 people from 10 for a committee: C(10,3) = 10!/(3! x 7!) = 120 ways. With replacement, use CR(n,r) = C(n+r-1, r). Combinations are used for lottery odds, team selection, and any scenario where arrangement order is irrelevant.
What is a Combination?
A combination is a selection of items from a collection where the order of selection does not matter. Unlike permutations, combinations treat ABC and CBA as the same selection. The combination formula C(n,r) = n!/(r!(n-r)!) counts the number of ways to choose r items from n items. Combinations are fundamental to probability theory, statistics, and are commonly used in lottery calculations, committee formation, and sampling problems.
Key Facts About Combinations
- Combination (nCr): order does NOT matter. Formula: C(n,r) = n!/(r!(n-r)!)
- C(n,r) = C(n, n-r) - choosing r items is same as leaving (n-r) items
- With replacement: CR(n,r) = C(n+r-1, r) allows repeated selections
- nCr is always less than or equal to nPr (permutations)
- Pascal's Triangle: each entry is C(row, position) = C(n,r)
- Sum of row n in Pascal's Triangle equals 2^n
- Lottery odds: C(49,6) = 13,983,816 combinations for 6 from 49
- nC0 = nCn = 1 (one way to choose nothing or everything)
Try These Examples
Quick-start with common scenarios
Practice Mode
Test your skills with practice problems
Practice with 5 problems to test your understanding.
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Quick Answer
Combinations count selections where order does NOT matter. Formula: C(n,r) = n! / (r! x (n-r)!). For example, choosing 3 people from 10 for a committee: C(10,3) = 10!/(3! x 7!) = 120 ways. With replacement, use CR(n,r) = C(n+r-1, r). Combinations are used for lottery odds, team selection, and any scenario where arrangement order is irrelevant.
Frequently Asked Questions
A combination is a selection of items where ORDER DOES NOT MATTER. C(n,r) counts ways to choose r items from n items. Example: Choosing 3 people from 10 for a committee = C(10,3) = 120 different groups.
Permutation: ORDER MATTERS (ABC differs from BAC). Use for rankings, arrangements. Combination: ORDER DOES NOT MATTER (ABC = BAC). Use for selections, committees. P(n,r) is always >= C(n,r).
Combinations with replacement allow selecting the same item multiple times. Formula: CR(n,r) = C(n+r-1, r). Example: Choosing 3 scoops of ice cream from 5 flavors (can repeat) = CR(5,3) = C(7,3) = 35.
Pascal's Triangle is a triangular array where each number is the sum of the two numbers above it. The entry at row n, position r equals C(n,r). It's useful for quickly finding combination values and has many mathematical properties.
Lottery odds use combinations because order does not matter. For a 6/49 lottery: C(49,6) = 49!/(6! x 43!) = 13,983,816 possible combinations. Your odds of winning are 1 in 13,983,816.
Last updated: 2025-01-15
Results
Combinations
120
C(10, 3)