Maths Extension 1 Year 11 - Module 4 - Lesson 4
Permutations with Restrictions
Name
Date
Class
1. Lesson focus
Permutations with restrictions. For adjacency, group the restricted objects as a block and account for their internal order; subtract forbidden cases from the total when useful.
Success criteria
- Apply the lesson method accurately in a routine example.
- Show each algebraic or domain step and check the result.
- Explain why the method and restrictions fit the question.
2. Worked practice
2. How many arrangements of ABCDE have A before B?
3. How many arrangements of 6 distinct people have A and B not adjacent?
3. Challenge
How many 5-digit even numbers can be made from 0,1,2,3,4,5 without repetition?
4. Multiple choice
For 6 distinct letters, treating a specified adjacent pair as one block gives how many linear arrangements?
A. 5!
B. 6!
C. 2×5!
D. 2×4!
Worked answer key
Compare each step with your working. Keep stated intervals, domains and restrictions in your final answer.
1. How many arrangements of ABCDE have A and B adjacent?
Treat AB/BA as one block: 4!×2=48.
2. How many arrangements of ABCDE have A before B?
By symmetry, half of 5! have A before B: 5!/2=60.
3. How many arrangements of 6 distinct people have A and B not adjacent?
Total 6!=720; adjacent cases are 5!×2=240; result 720−240=480.
Challenge. How many 5-digit even numbers can be made from 0,1,2,3,4,5 without repetition?
Last digit 0: 5×4!=120. Last digit 2 or 4: 2×4×4!=192. Total 312.
Multiple choice. For 6 distinct letters, treating a specified adjacent pair as one block gives how many linear arrangements?
Correct answer: C. The pair has two internal orders and the resulting five units can be arranged in 5! ways.
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