Permutations
Understand factorial notation, count ordered arrangements using nPr, and solve problems with restrictions.
What Are Permutations?
A permutation is an ordered arrangement of objects. Unlike combinations, the order in which items are selected matters. For example, the arrangements ABC and BAC are different permutations but would be the same combination.
Example: Arranging 3 Letters
How many ways can we arrange the letters A, B, C?
There are 3! = 3 × 2 × 1 = 6 permutations.
Factorial Notation
The factorial of a positive integer n, written as n!, is the product of all positive integers from 1 to n:
n! = n × (n-1) × (n-2) × ... × 2 × 1
By definition: 0! = 1
3! = 6
3 × 2 × 1
5! = 120
5 × 4 × 3 × 2 × 1
7! = 5040
7 × 6 × 5 × ... × 1
The nPr Formula and Restrictions
To choose and arrange r objects from n objects (where order matters), we use:
nPr = n! / (n - r)!
Read as "n permute r" -- the number of ways to arrange r items from n distinct items.
Restrictions add conditions that must be satisfied. Common restrictions include:
Fixed Position
A particular item must be in a specific position. Fix that item first, then arrange the rest.
Items Together
Treat the group as one "super item", arrange, then arrange within the group.
Items Not Together
Total arrangements minus arrangements where they ARE together.
Repeated Items
Divide by the factorial of the number of repeated items: n! / (p! × q! × ...)
Key Vocabulary
Permutation
An ordered arrangement of objects. The order matters -- ABC is different from BAC.
Factorial (n!)
The product of all positive integers from 1 to n. n! = n(n-1)(n-2)...1. Also, 0! = 1.
nPr
The number of permutations of r objects chosen from n distinct objects: nPr = n!/(n-r)!
Multiplication Principle
If task A can be done in m ways and task B in n ways, then both tasks together can be done in m × n ways.
Worked Examples
How many 3-letter "words" can be made from the letters A, B, C, D, E (no repetition)?
Step 1: We need to choose and arrange 3 letters from 5. This is 5P3.
Step 2: 5P3 = 5! / (5-3)! = 5! / 2! = 120/2 = 60.
Answer: 60 different 3-letter words.
In how many ways can 6 students line up if two specific students must stand together?
Step 1: Treat the two students as one "block". Now we have 5 items to arrange: 5! = 120.
Step 2: The two students within the block can swap: 2! = 2.
Step 3: Total = 5! × 2! = 120 × 2 = 240.
Answer: 240 arrangements.
How many distinct arrangements of the letters in MISSISSIPPI are there?
Step 1: Total letters: 11. Repeated: M(1), I(4), S(4), P(2).
Step 2: Arrangements = 11! / (1! × 4! × 4! × 2!) = 39916800 / (1 × 24 × 24 × 2).
Step 3: = 39916800 / 1152 = 34650.
Answer: 34,650 distinct arrangements.
Knowledge Check
Select the correct answer for each question. Click "Check Answer" to see if you are right.
Question 1
What is 6! ?
Question 2
Evaluate 8P3.
Question 3
How many ways can 4 books be arranged on a shelf?
Question 4
What is the value of 0! ?
Question 5
5 people sit in a row. If one particular person must sit at the left end, how many arrangements are there?
Key Concepts Summary
- ●A permutation is an ordered arrangement where order matters.
- ●n! (n factorial) counts the ways to arrange all n distinct objects.
- ●nPr = n!/(n-r)! counts the arrangements of r objects from n.
- ●For repeated items, divide by the factorials of the repetition counts.
- ●Handle restrictions by fixing constrained items first, or using complementary counting.