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Permutation Formula

Permutation Formula

A formula for the number of possible permutations of k objects from a set of n. This is usually written nPk .

 

Formula: Formula: nPk = n! / (n-k)! = n(n-1)(n-2)···(n-k+1)

Example:
How many ways can 4 students from a group of 15 be lined up for a photograph?
Answer:

There are 15P4 possible permutations of 4 students from a group of 15.

15P4 = 15!/11! = 15·14·13·12 = 32760, showing permutation of 4 students from 15. different lineups

 

See also

Combination formula, factorial

Key Formula

nPk=n!(n−k)!{}_{n}P_{k} = \frac{n!}{(n - k)!}
Where:
  • nn = The total number of distinct objects in the set
  • kk = The number of objects being chosen and arranged
  • n!n! = n factorial — the product of all positive integers from 1 to n
  • (n−k)!(n-k)! = The factorial of the difference, which cancels out the objects not selected

Worked Example

Problem: How many ways can 4 students from a group of 15 be lined up for a photograph?
Step 1: Identify n and k. There are 15 students total (n = 15) and you are choosing and arranging 4 of them (k = 4).
n=15,k=4n = 15, \quad k = 4
Step 2: Write the permutation formula and substitute the values.
15P4=15!(15−4)!=15!11!{}_{15}P_{4} = \frac{15!}{(15 - 4)!} = \frac{15!}{11!}
Step 3: Expand the factorial in the numerator only down to the point where it cancels with 11! in the denominator. This leaves the product of the top 4 factors of 15!.
15!11!=15×14×13×12\frac{15!}{11!} = 15 \times 14 \times 13 \times 12
Step 4: Multiply these values together.
15×14=210,210×13=2,730,2,730×12=32,76015 \times 14 = 210, \quad 210 \times 13 = 2{,}730, \quad 2{,}730 \times 12 = 32{,}760
Answer: There are 32,760 different ways to line up 4 students from a group of 15.

Another Example

This example shows permutations applied to assigning distinct roles rather than arranging people in a line, reinforcing that any situation where order or position matters calls for permutations.

Problem: A club has 8 members. In how many ways can a president, vice-president, and treasurer be chosen?
Step 1: Recognize this is a permutation problem because each position (president, vice-president, treasurer) is distinct — order matters. Here n = 8 and k = 3.
n=8,k=3n = 8, \quad k = 3
Step 2: Apply the permutation formula.
8P3=8!(8−3)!=8!5!{}_{8}P_{3} = \frac{8!}{(8 - 3)!} = \frac{8!}{5!}
Step 3: Cancel the common factorial and multiply the remaining factors.
8!5!=8×7×6=336\frac{8!}{5!} = 8 \times 7 \times 6 = 336
Answer: There are 336 ways to fill the three officer positions from 8 members.

Frequently Asked Questions

What is the difference between a permutation and a combination?
A permutation counts arrangements where the order matters (e.g., 1st place vs. 2nd place), while a combination counts selections where order does not matter (e.g., choosing a committee). The permutation formula nPk = n!/(n−k)! always gives a result equal to or larger than the combination formula nCk = n!/(k!(n−k)!), because each combination corresponds to multiple permutations.
When do you use the permutation formula?
Use the permutation formula whenever you are selecting k items from n items and the order of selection matters. Common scenarios include arranging people in a line, assigning ranked positions (1st, 2nd, 3rd), creating passwords or codes from a set of characters without repetition, and scheduling tasks in a specific sequence.
What is nPn equal to?
When k = n, the formula gives nPn = n!/(n−n)! = n!/0! = n!/1 = n!. This makes sense because arranging all n objects in a line is just n factorial. For example, the number of ways to arrange 5 books on a shelf is 5P5 = 5! = 120.

Permutation Formula vs. Combination Formula

Permutation FormulaCombination Formula
DefinitionNumber of ordered arrangements of k objects from nNumber of unordered selections of k objects from n
FormulanPk = n! / (n − k)!nCk = n! / (k!(n − k)!)
Does order matter?Yes — AB ≠ BANo — {A, B} = {B, A}
RelationshipnPk = nCk × k!nCk = nPk / k!
Example (n=5, k=3)5P3 = 605C3 = 10

Why It Matters

The permutation formula appears frequently in probability and statistics courses when you need to count favorable outcomes for events where arrangement matters. Standardized tests (SAT, ACT, AP Statistics) regularly include problems that require you to distinguish between permutations and combinations. Beyond the classroom, permutations model real-world problems like the number of possible PIN codes, race finish orders, or seating arrangements.

Common Mistakes

Mistake: Confusing permutations with combinations and forgetting to account for order.
Correction: Ask yourself: does rearranging the same items produce a different outcome? If choosing A then B is different from B then A, use the permutation formula. If they are the same, use the combination formula instead.
Mistake: Forgetting that 0! = 1, which causes errors when k = n.
Correction: Remember that 0! is defined as 1. So nPn = n!/0! = n!/1 = n!, not undefined.

Related Terms