TutorChase logo
CIE A-Level Maths Study Notes

4.2.3 Arrangements with Restrictions

The study of permutations and combinations in mathematics offers a multitude of intriguing problems, particularly when it comes to arranging objects with certain restrictions.

Restricted Arrangements

  • Objective: To arrange items with specific rules or limitations.
  • Application: Useful in complex problems, especially in combinatorics.

Key Concepts

1. Fixed Position: Some items must be in specific places.

2. Adjacency: Certain items must or must not be next to each other.

3. Separation: Specific spacing between items is required.

Strategy for Solving

  • Total Minus Restricted: First calculate all possible arrangements, then subtract those that break the rules.

Examples

Example 1. Arranging People in a Line

People in line

Image courtesy of Vertor Stock

Objective: Arrange Alice, Ben, Carol, Dave, and Emma in a line with the restriction that Alice and Ben are not next to each other.

Solution:

  • Total Unrestricted Arrangements:
    • 5!=1205! = 120
  • Restricted Arrangements (Alice and Ben together):
    • Treat Alice and Ben as one unit: 4!4! arrangements
    • Alice and Ben can switch places: 2!2!
    • Total restricted arrangements: 24×2=4824 \times 2 = 48
  • Final Calculation:
    • Subtract restricted from total: 12048=72120 - 48 = 72

Result: There are 72 ways to arrange these five people with Alice and Ben not next to each other.

Example 2. Arranging Letters in 'MATHEMATICS'

Mathematics

Image courtesy of Brainly

Objective: Arrange the letters in "MATHEMATICS" such that the two 'M's are never adjacent.

Solution:

  • Total Arrangements Without Restriction:
    • Formula: 11!2!×2!×2!\frac{11!}{2! \times 2! \times 2!}
    • Calculation: 399168008=4989600\frac{39916800}{8} = 4989600
  • Arrangements with M's Adjacent:
    • Treat 'MM' as one unit, reducing to "MMATHEATICS".
    • Formula: (10!2!×2!)×2!\left( \frac{10!}{2! \times 2!} \right) \times 2!
    • Calculation: (36288004)×2=1814400\left( \frac{3628800}{4} \right) \times 2 = 1814400
  • Final Calculation:
    • Subtract arrangements with 'M's adjacent from total.
    • Calculation: 49896001814400=31752004989600 - 1814400 = 3175200

Result: There are 3,175,200 distinct arrangements of "MATHEMATICS" where the two 'M's are not adjacent.

Example 3: Multi-row Arrangements

Objective: Arrange six students in two rows of three, with restrictions on placements.

Solution:

  • Each Row Arrangements: 3!=63! = 6
  • Restrictions (A and B in Same Row):
    • 2×2!=42 \times 2! = 4 (for A and B together in one row)
  • Final Calculation:
    • Total without restriction: 6×6=366 \times 6 = 36
    • Restricted (A and B together): 4×2=84 \times 2 = 8
    • Total with restriction: 368=2836 - 8 = 28

Result: 2828 distinct arrangements.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

Hire a tutor

Please fill out the form and we'll find a tutor for you.

1/2 About yourself
Still have questions?
Let's get in touch.