Mathwords logoReference LibraryMathwords

Permutation

Permutation

A selection of objects in which the order of the objects matters.

 

Example: The permutations of the letters in the set {a, b, c} are:

abc        acb
bac        bca
cab        cba

 

 

See also

Permutation formula, combination

Key Formula

P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n - r)!}
Where:
  • nn = Total number of objects available
  • rr = Number of objects you are choosing and arranging
  • n!n! = n factorial — the product of all positive integers from 1 to n

Worked Example

Problem: A club has 10 members. In how many ways can they choose a president, vice-president, and treasurer?
Identify n and r: There are 10 members to choose from (n = 10), and you are filling 3 distinct positions (r = 3). Because each position is different, order matters — this is a permutation.
Apply the formula: Substitute into the permutation formula.
P(10,3)=10!(10−3)!=10!7!P(10, 3) = \frac{10!}{(10 - 3)!} = \frac{10!}{7!}
Simplify: Most of the factorial terms cancel. You only need the first three factors of 10!.
10!7!=10×9×8=720\frac{10!}{7!} = 10 \times 9 \times 8 = 720
Answer: There are 720 different ways to choose a president, vice-president, and treasurer from 10 members.

Another Example

Problem: How many different 4-letter arrangements can be formed from the letters A, B, C, D, E, F (no repeats)?
Set up: You have 6 letters (n = 6) and want to arrange 4 of them (r = 4). Since rearranging the same 4 letters produces a different arrangement, order matters.
P(6,4)=6!(6−4)!=6!2!P(6, 4) = \frac{6!}{(6 - 4)!} = \frac{6!}{2!}
Calculate: Expand and simplify.
6!2!=7202=360\frac{6!}{2!} = \frac{720}{2} = 360
Answer: There are 360 possible 4-letter arrangements.

Frequently Asked Questions

What is the difference between a permutation and a combination?
A permutation counts arrangements where order matters; a combination counts selections where order does not matter. For example, choosing president then vice-president is a permutation problem, while choosing a 2-person committee (where no one has a special role) is a combination problem. Permutations always produce a larger count than combinations for the same n and r (when r > 1).
When do I use permutations instead of combinations?
Use permutations when the positions or slots you are filling are distinct — rankings, sequences, assigned roles, PIN codes, or any situation where rearranging the same items gives a different outcome. If swapping two chosen items changes nothing (like picking team members with no assigned positions), use combinations instead.

Permutation vs. Combination

PermutationCombination
Key DifferenceCounts ordered arrangementsCounts unordered selections
FormulaP(n,r)=n!(n−r)!P(n,r) = \frac{n!}{(n-r)!}C(n,r)=n!r!(n−r)!C(n,r) = \frac{n!}{r!(n-r)!}
Example (choose 3 from 5)P(5,3)=60P(5,3) = 60C(5,3)=10C(5,3) = 10
RelationshipAlways larger — each combination has r!r! permutationsC(n,r)=P(n,r)r!C(n,r) = \frac{P(n,r)}{r!}

Why It Matters

Permutations appear whenever you need to count ordered outcomes — passwords, race finishes, seating charts, or scheduling. They form the foundation of counting principles in probability, since knowing the total number of ordered outcomes lets you calculate how likely a specific arrangement is. Many standardized tests and real-world problems in cryptography, logistics, and computer science rely on permutation counting.

Common Mistakes

Mistake: Using the permutation formula when order does not matter.
Correction: Always ask: does rearranging the chosen items create a different outcome? If not, you need the combination formula, which divides out the extra arrangements by r!.
Mistake: Confusing n and r — for example, setting r as the total and n as the number chosen.
Correction: Remember that n is always the larger number (the total pool) and r is the number you select from it. P(n, r) requires n ≥ r.

Related Terms