Variations Calculator

Enter n and k. The variation V(n,k) = n!/(n−k)! counts ordered arrangements of k distinct items from n. V(5,3) = 60; V(10,3) = 720. Order matters, no repetition — usually written P(n,k) or k-permutations. This is not combinations C(n,k) and not variations with repetition n^k.

Input data

Result

Enter n and k.

Variations without repetition. V(n,k) = n!/(n−k)!. When k > n the result is 0. V(n,n) = n!.

How it works

How V(5,3) is computed
n5
k3
n!120
(n−k)!2
V(n,k)60

A variation without repetition V(n,k), usually written P(n,k) or “k-permutation” in English texts, counts ways to line up k distinct items from an n-element set when order matters. For n = 5 and k = 3 the result is 60: from five runners on a three-place podium, Anna–Bartek–Celina differs from Celina–Bartek–Anna.

The formula is V(n,k) = n! / (n−k)! for 0 ≤ k ≤ n. Equivalently the falling product n·(n−1)·…·(n−k+1). When k > n the result is 0. V(5,3) = 120 / 2 = 60. V(6,2) = 720 / 24 = 30. Link to combinations: V(n,k) = C(n,k)·k!.

The calculator uses the numbers you enter in fields n (wariacje-a) and k (wariacje-b). It takes floor. For negative n or k it reports an error. It computes via the falling product and the panel also shows n!, (n−k)!, and C(n,k).

The result panel shows V(n,k), echoes of n and k, n!, (n−k)!, the falling product, and C(n,k). For n ≤ 10 a bar chart draws V(n,0)…V(n,n). A seat diagram (for moderate n, k) illustrates the successive factors n, n−1, …. Examples: V(5,3) = 60; V(6,2) = 30; V(4,4) = 24; V(5,1) = 5; V(8,0) = 1; V(7,2) = 42; V(10,3) = 720; when k > n — 0.

A separate combinations page counts unordered C(n,k). Permutations P(n) = n! arrange the whole set. This calculator stays with ordered V(n,k) without repetition — it does not list tuples and does not count variations with repetition n^k.

Formula

V(n,k) = P(n,k) = n! / (n−k)! for 0 ≤ k ≤ n; when k > n → 0.

Equivalently: V(n,k) = n · (n−1) · … · (n−k+1).

Bridge to combinations: V(n,k) = C(n,k) · k!.

Symbols: n — set size (the “n” field); k — arrangement length (the “k” field). Mathematical model: k-permutation / variation without repetition (Brualdi; Concrete Mathematics).

Theory, falling product, and the link to C(n,k)

Definition

Let n and k be non-negative integers. A variation without repetition V(n,k) — in English combinatorics usually P(n,k), a “k-permutation of n”, or the number of injections from a k-set into an n-set — is the number of ordered k-tuples of pairwise distinct entries chosen from an n-element set. The classical formula is V(n,k) = n! / (n−k)! for 0 ≤ k ≤ n. Equivalently the falling factorial n^{\underline{k}} = n·(n−1)·…·(n−k+1). When k > n no such tuple exists without repetition, so this calculator returns 0.

Polish school texts prefer “wariacja”; English sources avoid “variation” as a calque and write P(n,k). MathWorld (Permutation) and textbooks by Brualdi and Concrete Mathematics (Graham, Knuth, Patashnik) treat P(n,k) as the injection count or as the number of ways to fill k ordered seats from n candidates without reuse.

This page does not divide full factorials for large n directly: it computes ∏_{i=0}^{k−1} (n − i) and also shows n! and (n−k)! in the panel as a check of the quotient form. For typical school data the result matches the definition.

Historical and conceptual context

The multiplication rule (“n choices for the first seat, n−1 for the second…”) appears in medieval and early modern texts long before modern factorial notation. Christian Kramp (1760–1826) popularized the symbol n! for 1·2·…·n. Nineteenth-century enumerative combinatorics separated full orders (permutations), partial orders (k-permutations / variations), and unordered selections (combinations).

Conceptually, variations sit between permutations and combinations. Permutations P(n) = n! arrange all items; combinations C(n,k) = n!/(k!(n−k)!) pick a set without order; variations V(n,k) = n!/(n−k)! arrange k of n with order. Bridge: V(n,k) = C(n,k)·k!. In computer science P(n,k) counts injections, codes without repeated symbols, and drives partial-permutation generators. In probability it appears in ordered sampling without replacement.

This calculator does not solve for unknown n or k, does not list tuples, and does not count variations with repetition (that formula is n^k). It answers one question: how many ordered picks of k from n without repetition.

Formal properties

Domain on this page: n, k ∈ ℕ₀. For 0 ≤ k ≤ n: V(n,k) = n!/(n−k)! = n^{\underline{k}}. V(n,0) = 1. V(n,1) = n. V(n,n) = n!. V(n,k) = 0 when k > n. Recurrence: V(n,k) = n · V(n−1,k−1) for k ≥ 1 (with V(m,0) = 1). Alternatively V(n,k) = (n−k+1) · V(n,k−1) for k ≥ 1.

Link to combinations: V(n,k) = C(n,k)·k! and C(n,k) = V(n,k)/k!. Link to full permutations: V(n,n) = P(n) = n!. For fixed n ≥ 1 the sequence in k usually rises to V(n,n) = n!.

The quotient identity V(n,k)/V(n,k−1) = n−k+1 for 1 ≤ k ≤ n ties neighboring lengths. Variations with repetition n^k do not divide by the factorial of unused items — a different model.

Full derivation

Step 1 (multiplication rule). The first of k ordered seats has n choices, the second has n−1 remaining, …, the k-th has n−k+1. The product n·(n−1)·…·(n−k+1) is V(n,k).

Step 2 (factorials). Full factorial n! counts all orders of n items. Orders of the unused n−k items are (n−k)!. Dividing n! by (n−k)! strips the orders you never arrange: V(n,k) = n!/(n−k)!.

Step 3 (bridge to C). Each unordered k-set corresponds to exactly k! orderings. Hence C(n,k) = V(n,k)/k! and V(n,k) = C(n,k)·k!.

Step 4 (boundaries). For k = 0 the empty product is 1. For k = n the denominator is 0! = 1 and V(n,n) = n!. For k > n the falling product has no meaning without repetition — result 0.

Step 5 (examples). V(5,3) = 5·4·3 = 60 = 120/2. V(6,2) = 6·5 = 30. V(4,4) = 24. V(3,5) = 0.

Special cases

V(n,2) = n(n−1) — ordered pairs of distinct items. A top-3 ranking from 20 athletes: V(20,3) = 20·19·18 = 6,840. A 4-letter code from 26 distinct letters: V(26,4) = 26·25·24·23 = 358,800 (not the with-repetition model).

Negative n or k after flooring are rejected with an error (undefined). Non-integers are floored: 5.8 and 3.2 give V(5,3) = 60. When k > n → 0 (not a form error).

For very large n, floating-point arithmetic loses integer exactness; larger n needs big-integer arithmetic. This educational calculator targets the school range and the panel examples.

Numerical remarks

The falling product limits magnitude growth versus a direct n! / (n−k)! split for large n and small k. The panel still shows n! and (n−k)! as an educational breakdown — for large n those factorials may become Infinity while V stays finite. Comma and period are equivalent.

The bar chart is limited to n ≤ 10, and the seat diagram to moderate n and k, so graphics stay readable. C(n,k) in the panel is the check V(n,k)/k!, not a separate form mode.

Numbers are not sent to a server; the browser computes everything in floating-point arithmetic.

Related notions and sources

Related: factorial n!, permutations P(n), combinations C(n,k), variations with repetition n^k, falling factorial, finite injections, codes without repeated symbols. The identity V(n,k) = C(n,k)·k! ties the three classic school counts.

Sources: Brualdi, Introductory Combinatorics (permutations / P(n,k)); Graham, Knuth, Patashnik, Concrete Mathematics (falling factorials); MathWorld — Permutation; Wikipedia — Permutation / k-permutation. Educational tool — not a substitute for a textbook or teacher check.

How to use

  1. Enter n (set size), for example 5.
  2. Enter k (arrangement length), for example 3.
  3. Click Calculate or wait for the live result. V(5,3) = 60. V(6,2) = 30.
  4. Check the panel: n!, (n−k)!, falling product, and C(n,k).
  5. V(4,4) = 24. V(8,0) = 1. When k > n the result is 0. Unordered picks live on combinations.

Symbol glossary

V(n,k) = n!/(n−k)!. V(5,3) = 60. V(6,2) = 30. Order matters.

V(n,k)
Number of ordered k-tuples from n. V(5,3) = 60. Also written P(n,k).
n
Set size in the “n” field. When k > n the result is 0.
k
Arrangement length in the “k” field. k = n returns to factorial. k = 0 yields 1.
n!
Numerator of the quotient form. V(n,n) = n!.
(n−k)!
Denominator — factorial of unused items. At k = n it is 0! = 1.
C(n,k)
Combinations: C(n,k) = V(n,k)/k!. For (5,3) it is 10.

Examples

Example 1

  • n = 5
  • k = 3

60

What is V(5,3)? 60 (5·4·3).

Example 2

  • n = 6
  • k = 2

30

What is V(6,2)? 30 (6·5).

Example 3

  • n = 4
  • k = 4

24

What is V(4,4)? 24 = 4!.

Example 4

  • n = 5
  • k = 1

5

What is V(5,1)? 5.

Example 5

  • n = 8
  • k = 0

1

What is V(8,0)? 1 (empty list).

Example 6

  • n = 7
  • k = 2

42

What is V(7,2)? 42; C(7,2) = 21.

Example 7

  • n = 10
  • k = 3

720

What is V(10,3)? 720 (10·9·8).

Example 8

  • n = 3
  • k = 5

0

What is V(3,5)? 0 (k > n).

Related calculators

Common questions

What is the domain of V(n,k) on this page?

After flooring, n and k are non-negative integers. For 0 ≤ k ≤ n the result is n!/(n−k)!. When k > n the result is 0. Negative n or k leave the value undefined — the calculator reports an error.

What does variation without repetition mean?

You line up k distinct items from an n-element set and do not reuse an item. Order makes a new result: podium Anna–Bartek–Celina differs from Celina–Bartek–Anna. That is V(n,k), usually written P(n,k) or “k-permutations” in English texts.

How does V(n,k) differ from C(n,k) and from P(n)?

C(n,k) = n!/(k!(n−k)!) chooses a set without order. Link: V(n,k) = C(n,k)·k!. Arrangements of the whole set are P(n) = n! = V(n,n) on a separate page. For n = 5 and k = 2: V(5,2) = 20 while C(5,2) = 10.

Is V(n,n) a factorial?

Yes. V(4,4) = 24 = 4!. When you seat everyone, you are left with the permutations of n items. The panel also shows n! and (n−n)! = 0! = 1.

What if k is larger than n?

The result is 0. You cannot line up more distinct items than the set holds. That differs from the combinations page, which reports an error instead of zero when k > n.

What are V(n,0) and V(5,1)?

V(n,0) = 1 — one empty list (denominator (n−0)! = n!). V(5,1) = 5 — one seat from five is simply n.

How does the calculator compute the number?

Instead of dividing huge factorials directly, it uses the falling product n·(n−1)·…·(n−k+1), equal to n!/(n−k)!. For a legal n and k the value is an integer. The panel also shows n!, (n−k)!, and C(n,k).

Do the currency and unit settings in the page header affect the result?

No. The result depends only on n and k. The currency and unit settings in the page header do not affect this result.

Do commas in the fields work?

Yes. Flooring uses floor: 5.8 and 3.2 compute as V(5,3) = 60. Comma and period are equivalent in the parser.

Where do I compute combinations or full-set permutations?

Unordered picks — on the combinations C(n,k) page. Arrangements of all n items — on permutations P(n) = n! or factorial. Here only ordered V(n,k) without repetition.

Is this variations with repetition?

No. Variations with repetition count n^k (each position may reuse an item). This calculator counts only variations without repetition: n!/(n−k)!.

What if one field is empty?

Empty n or empty k leaves the result blank. The calculator does not substitute zero for a missing entry. Enter zero on purpose when needed (for example V(8,0) = 1).

How do I calculate variations without repetition?

Use V(n,k) = n!/(n−k)! or the falling product n·(n−1)·…·(n−k+1). For n = 5 and k = 3: 5·4·3 = 60.

What is V(5,3)?

60. That is the first example on this page: 5·4·3 = 60 podium arrangements from five people.

What is V(10,3)?

720. Seventh example: 10·9·8 = 720. In English texts that is the same as P(10,3).

What is V(8,3)?

336. V(8,3) = 8·7·6 = 336 — ways to award three medals to eight athletes.

What is V(10,4)?

5040. V(10,4) = 10·9·8·7 = 5040. That is not 10^4 = 10,000 (variations with repetition).

The result is V(n,k): for n = 5 and k = 3 it is 60. V(n,n) = n!. Unordered picks live on combinations.

Knowledge sources

Definition, falling product, and the link to C(n,k) below come from established sources.

Page updated November 3, 2026.