Example 1
- n = 5
- k = 3
60
What is V(5,3)? 60 (5·4·3).
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.
Enter n and k.
Variations without repetition. V(n,k) = n!/(n−k)!. When k > n the result is 0. V(n,n) = n!.
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.
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).
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.
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.
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.
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.
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.
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: 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.
V(n,k) = n!/(n−k)!. V(5,3) = 60. V(6,2) = 30. Order matters.
60
What is V(5,3)? 60 (5·4·3).
30
What is V(6,2)? 30 (6·5).
24
What is V(4,4)? 24 = 4!.
5
What is V(5,1)? 5.
1
What is V(8,0)? 1 (empty list).
42
What is V(7,2)? 42; C(7,2) = 21.
720
What is V(10,3)? 720 (10·9·8).
0
What is V(3,5)? 0 (k > n).
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.
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.
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.
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.
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.
V(n,0) = 1 — one empty list (denominator (n−0)! = n!). V(5,1) = 5 — one seat from five is simply n.
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).
No. The result depends only on n and k. The currency and unit settings in the page header do not affect this result.
Yes. Flooring uses floor: 5.8 and 3.2 compute as V(5,3) = 60. Comma and period are equivalent in the parser.
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.
No. Variations with repetition count n^k (each position may reuse an item). This calculator counts only variations without repetition: n!/(n−k)!.
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).
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.
60. That is the first example on this page: 5·4·3 = 60 podium arrangements from five people.
720. Seventh example: 10·9·8 = 720. In English texts that is the same as P(10,3).
336. V(8,3) = 8·7·6 = 336 — ways to award three medals to eight athletes.
5040. V(10,4) = 10·9·8·7 = 5040. That is not 10^4 = 10,000 (variations with repetition).
Definition, falling product, and the link to C(n,k) below come from established sources.
Page updated November 3, 2026.