Combinations without Repetition

Calculate possible combinations from a set (Binomial coefficient)

Binomial coefficient "n choose k": Number of ways to choose k objects from n

Combinations Calculator

Combinations without Repetition

Calculates C(n,k) - the number of ways to choose k objects from n, without regard to order and without repetition.

Total number of available objects
Number of objects to be selected (k ? n)
Calculation Result
C(n,k) =

Combination Example

Default Example: C(8,3)
Total set: n = 8
Selection: k = 3
Result: C(8,3) = 56
56 different 3-element groups from 8 objects
Simple Example: 3 from 5

Objects: A, B, C, D, E

Possible 3-element combinations:

ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE

C(5,3) = 10 combinations

Important Properties
  • Order is irrelevant: ABC = BAC = CAB
  • No repetition: Each object max. once
  • C(n,k) = C(n,n-k) (Symmetry)
  • C(n,0) = C(n,n) = 1

Mathematical Foundations of Combinatorics

Combinations without repetition are based on the binomial coefficient and factorial:

Binomial Coefficient
\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]

"n choose k" - Number of k-subsets of an n-set

Factorial Definition
\[n! = n \times (n-1) \times \ldots \times 2 \times 1\]

With the convention: 0! = 1

Combinatorics Formulas and Examples

General Combination Formula
\[C(n,k) = \binom{n}{k} = \frac{n!}{k! \cdot (n-k)!}\]

Number of ways to choose k objects from n (without regard to order)

Step-by-Step Calculation: C(8,3)

Given: n = 8, k = 3

1. Substitute in formula:

\[C(8,3) = \frac{8!}{3! \cdot (8-3)!} = \frac{8!}{3! \cdot 5!}\]

2. Calculate factorials:

8! = 8 × 7 × 6 × 5! = 40320

3! = 3 × 2 × 1 = 6

5! = 120

3. Simplification:

\[C(8,3) = \frac{8 \times 7 \times 6 \times 5!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3!} = \frac{336}{6} = 56\]
Additional Calculation Examples
Calculate C(5,2):
\[C(5,2) = \frac{5!}{2! \cdot 3!} = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10\]
Calculate C(10,4):
\[C(10,4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210\]
C(6,0) and C(6,6):
\[C(6,0) = C(6,6) = 1\]
Symmetry Property:
\[C(n,k) = C(n,n-k)\]
Pascal's Triangle
n=0:               1
n=1:            1   1
n=2:          1   2   1
n=3:        1   3   3   1
n=4:      1   4   6   4   1
n=5:    1   5  10  10   5   1
n=6:  1   6  15  20  15   6   1

Each number is C(n,k) for the corresponding row n and position k

Combinatorics Reference

Default Example
C(8,3) = 56 8! / (3! × 5!) 336 / 6 = 56
Special Values

C(n,0) = 1: Empty set

C(n,1) = n: Single selection

C(n,2) = n(n-1)/2: Pairs

C(n,n) = 1: Select all

Important Properties

Symmetry: C(n,k) = C(n,n-k)

Pascal Identity: C(n,k) = C(n-1,k-1) + C(n-1,k)

Sum: ?C(n,k) = 2?

Alternating: ?(-1)?C(n,k) = 0

Applications

Lottery: C(49,6) = 13,983,816

Team formation: From 12 people

Probability: Favorable cases

Urn model: Without replacement

Combinations without Repetition - Detailed Description

Fundamentals of Combinatorics

Combinations without repetition are a fundamental concept in combinatorics. They answer the question: "In how many ways can I choose k objects from n different objects, when the order is not important and each object can be selected at most once?"

Characteristics:
• Order is irrelevant: {A,B,C} = {C,A,B}
• No repetition: Each element max. once
• Subsets of a base set
• Binomial coefficient as mathematical representation

Calculation Methods

The calculation is done via the binomial coefficient, which is also read as "n choose k". The formula is based on factorials and can be efficiently calculated for large numbers through clever simplification.

Efficient Calculation

Instead of calculating all factorials: C(n,k) = (n×(n-1)×...×(n-k+1)) / k!
Use symmetry: C(n,k) = C(n,n-k) for k > n/2

Practical Applications

Combinations without repetition find applications in many areas: from probability calculations to optimization and everyday decisions like team formation or menu selection.

Typical Scenarios:
• Lottery probabilities
• Team compositions
• Selection problems
• Statistical sampling

Pascal's Triangle

Pascal's triangle is an elegant representation of all binomial coefficients. Each number is created by adding the two numbers above it, illustrating the Pascal identity C(n,k) = C(n-1,k-1) + C(n-1,k).

Pascal Identity

The recursive relationship: "To choose k from n, I can either take the first element and choose k-1 from the remaining n-1, or leave out the first element and choose k from the remaining n-1."

Practical Examples and Applications

Lottery 6 from 49

Problem: Probability for jackpot

Calculation: C(49,6)

Result: 13,983,816 possibilities

Chance: 1:14 million

Team Formation

Problem: 5-person team from 12 people

Calculation: C(12,5)

Result: 792 possibilities

Benefit: Fair team rotation

Menu Selection

Problem: Choose 3 dishes from 8

Calculation: C(8,3)

Result: 56 combinations

Benefit: Plan menu variety

Combinatorics vs. Other Counting Principles
  • Combinations without rep.: C(n,k) - Order irrelevant, no rep.
  • Combinations with rep.: C(n+k-1,k) - Order irrelevant, with rep.
  • Permutations without rep.: P(n,k) = n!/(n-k)! - Order matters
  • Variations with rep.: n? - Order matters, with rep.
  • Urn model: Drawing without replacement, order irrelevant
  • Binomial theorem: (a+b)? = ?C(n,k)a???b?


Other Combinatorics Functions

Combinations with Repetition  •  Combinations without Repetition  •  Permutations  •  Rule of Product  •  Variations with Repetition  •  Variations without Repetition  •  Activity Selection Problem  •