Variations without Repetition
Calculate ordered subsets - k-permutations from n elements
Variation V(n,k): Number of ordered k-selections from n elements
Variations Calculator
Variations without Repetition
Calculates V(n,k) - the number of ways to select k objects from n and arrange them in different orders.
Variations Example
Default Example: V(8,3)
Concrete Example: 2 from {A, B, C}
Objects: A, B, C → select 2 and arrange:
V(3,2) = 6 variations
Important Properties
- Order is crucial: AB ≠ BA
- Partial selection: k ≤ n (not all objects)
- Without repetition: Each object at most once
- V(n,k) = n!/(n-k)! (Partial factorial)
Mathematical Foundations of k-Permutations
Variations without repetition are based on partial factorials and the permutation principle:
k-Permutation
Ordered selection of k objects from n different ones
Step-by-step Multiplication
k consecutive factors starting from n downward
Variation Formulas and Examples
General Variation Formula
Ordered selection of k objects from n different ones (k ≤ n)
Step-by-Step Calculation: V(8,3)
Given: n = 8 objects, k = 3 to select
1. Apply factorial formula:
\[V(8,3) = \frac{8!}{(8-3)!} = \frac{8!}{5!}\]2. Cancel factorials:
\[V(8,3) = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6\]3. Step-by-step multiplication:
8 × 7 = 56
56 × 6 = 336
4. Result:
\[V(8,3) = 336 \text{ different arrangements}\]Intuitive Explanation with Positions
8 objects {A, B, C, D, E, F, G, H}, fill 3 positions:
Position 1: 8 choices (all objects available)
Position 2: 7 choices (1 already chosen)
Position 3: 6 choices (2 already chosen)
Total: 8 × 7 × 6 = 336 different 3-element arrangements
Additional Calculation Examples
Calculate V(5,2):
Calculate V(10,4):
V(n,1) - Special case:
V(n,n) - Special case:
Complete Example: V(3,2)
Objects: {A, B, C}, select 2 and arrange
Calculation:
\[V(3,2) = \frac{3!}{1!} = 3 \times 2 = 6\]All variations:
6 different ordered 2-element selections from 3 objects
Comparison: Variations vs. Other Counting Principles
Variations V(n,k)
Select k from n
Order matters
Without repetition
V(3,2) = 6
Combinations C(n,k)
Select k from n
Order doesn't matter
Without repetition
C(3,2) = 3
Permutations P(n)
Use all n
Order matters
Without repetition
P(3) = 6
Relation: V(n,k) = C(n,k) × k! and V(n,n) = P(n) = n!
Variations Reference
Default Example
Special Values
V(n,0) = 1: Empty selection
V(n,1) = n: Single selection
V(n,n) = n!: Use all = Permutation
V(n,k) ≥ C(n,k): Always greater or equal
Properties
Order: AB ≠ BA ≠ ...
Partial selection: k ≤ n objects
Without repetition: Each at most once
k factors: n×(n-1)×...×(n-k+1)
Applications
Races: Placements (1st, 2nd, 3rd)
Codes: Position-wise without repetition
Awards: Different prizes
Scheduling: Priority ordering
Variations without Repetition - Detailed Description
Understanding k-Permutations
Variations without repetition are ordered subsets and form the bridge between combinations and permutations. They answer the question: "In how many ways can I select and arrange k different objects from n?"
• Partial selection: k ≤ n (not all objects)
• Order matters: AB ≠ BA
• Without repetition: Each element at most once
• Partial factorial: n!/(n-k)!
Partial Factorials
The formula V(n,k) = n!/(n-k)! can be understood as a "partial factorial": Take the first k factors of factorial n! and omit the last (n-k) factors by dividing by (n-k)!.
Cancellation Rule
n!/(n-k)! = n×(n-1)×(n-2)×...×(n-k+1)
Exactly k consecutive factors starting from n downward
Practical Applications
Variations are relevant everywhere both selection and arrangement play a role: competition rankings, passwords without repetition, task assignment with priorities.
• Podium placements (1st, 2nd, 3rd place)
• Leadership positions in organizations
• PIN codes without repeated digits
• Starting lineup in races
Relationship to Other Counting Principles
Variations are directly related to combinations: V(n,k) = C(n,k) × k!. This means: First choose (combination), then arrange (permutation of the selection).
Relation to Combinations
A variation is a combination multiplied by all possible arrangements of that combination: V(n,k) = C(n,k) × k!
Practical Examples and Applications
Race Podium
Problem: 10 runners, top 3 places
Calculation: V(10,3)
Result: 10×9×8 = 720 possibilities
Meaning: 720 different podium arrangements
Leadership Election
Problem: 12 candidates, 4 positions
Calculation: V(12,4)
Result: 12×11×10×9 = 11,880
Meaning: Different position assignments
PIN without Repetition
Problem: 4-digit PIN, 10 digits
Calculation: V(10,4)
Result: 10×9×8×7 = 5,040
Meaning: PINs without repeated digits
Variants of the Variation Principle
- Without repetition: V(n,k) = n!/(n-k)! - Each element at most once
- With repetition: n^k - Elements can be used multiple times
- Circular variations: Arrangements on a circle
- Restricted variations: With conditions and exclusions
- Multiset variations: With multiple instances of object types
- Rank variations: Hierarchical arrangements
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