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Permutation Calculator -- nPr

Calculate P(n, r) = n!/(n-r)! -- the number of ordered arrangements of r items from a set of n.

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Die Gesamtzahl der verschiedenen Elemente in der Menge

Die Anzahl der aus der Gesamtmenge gewählten oder angeordneten Elemente

Formel
P(n, r)=n!(n−r)!P(n,\,r) = \dfrac{n!}{(n-r)!}

Beispiel ausprobieren

Wähle ein Szenario, um zu sehen, wie der Rechner funktioniert, und passe dann die Werte an

Race Podium — Top 3 from 10 Runners

How many different 1st/2nd/3rd-place finishes can come from a 10-person race?

Wichtige Werte: n = 10 · r = 3 · P(10,3) = 720

4-Digit PIN from 10 Digits (No Repetition)

A lock allows digits 0–9 with no repeated digits. How many PINs are possible?

Wichtige Werte: n = 10 · r = 4 · P(10,4) = 5,040

Dokumentation

What Is a Permutation?

A permutation is an ordered arrangement of objects. The number of ways to arrange rr items chosen from nn distinct items is:

P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}

The key word is order matters. The arrangement ABC is different from BCA — both count as separate permutations.

Permutations vs. Combinations

FeaturePermutation P(n, r)Combination C(n, r)
Order matters?YesNo
Formulan!(n−r)!\frac{n!}{(n-r)!}n!r!(n−r)!\frac{n!}{r!(n-r)!}
RelationshipP(n,r)=r!×C(n,r)P(n,r) = r! \times C(n,r)
Example: 3 from 56010
Use caseRankings, codes, sequencesTeams, committees, selections

Memory aid: “Does rearranging change the outcome?” If a gold-silver-bronze podium is different from silver-gold-bronze, use permutations. If picking 3 team members is the same regardless of order, use combinations.

Permutations with Repetition

When items can be reused (like digits in a PIN), the count is simpler:

nrn^r
ScenarionrCount
4-digit PIN (0–9)10410,000
Binary string of length 828256
3-letter codes (A–Z)26317,576
License plate (3 letters + 4 digits)26³ × 10⁴—175,760,000

Permutations of a Multiset

When some items are identical, divide out the repeated arrangements:

n!n1!⋅n2!⋅…⋅nk!\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}

Example: How many distinct arrangements of the letters in MISSISSIPPI?

11!1!⋅4!⋅4!⋅2!=39,916,8001⋅24⋅24⋅2=34,650\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39{,}916{,}800}{1 \cdot 24 \cdot 24 \cdot 2} = 34{,}650

Frequently Asked Questions

What is a permutation?

A permutation is an ordered arrangement of objects. The number of ways to arrange rr items from nn distinct items is P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}. Unlike combinations, order matters: ABC and BAC count as different permutations.

How do you calculate P(n, r)?

P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}. For example, P(5,3)=5!2!=1202=60P(5, 3) = \frac{5!}{2!} = \frac{120}{2} = 60. This counts the number of ordered sequences of 3 items chosen from 5 distinct items.

What is the relationship between permutations and combinations?

P(n,r)=r!×C(n,r)P(n, r) = r! \times C(n, r). Permutations count all ordered arrangements, while combinations count unordered selections. Dividing permutations by r!r! removes the ordering to give combinations.

What are permutations with repetition?

When items can be reused (like digits in a PIN), the count is nrn^r. For example, a 4-digit PIN using digits 0--9 has 104=10,00010^4 = 10{,}000 possible arrangements. Each position has nn choices, independent of previous selections.

How do you count permutations of a word with repeated letters?

Divide the total factorial by the factorials of each repeated letter. For MISSISSIPPI (11 letters: 1 M, 4 I, 4 S, 2 P), the count is 11!1!×4!×4!×2!=34,650\frac{11!}{1! \times 4! \times 4! \times 2!} = 34{,}650 distinct arrangements.

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