ZIMSEC A Level · Statistics · Form 5
Arrangements
- n distinct objects can be arranged in n! ways.
- With repeated objects: n!/(p! q! …).
- Permutations of r from n: ⁿPᵣ = n!/(n − r)!.
- Combinations (order does not matter): ⁿCᵣ = n!/(r!(n − r)!).
Worked example: How many arrangements of the letters of ZIMBABWE are there?
8 letters with B repeated twice → 8!/2! = 40 320/2 = 20 160
Worked example: A committee of 4 is chosen from 6 boys and 5 girls. How many committees have exactly 2 girls?
⁵C₂ × ⁶C₂ = 10 × 15 = 150
Restrictions
- Items together: treat them as one block, then multiply by their internal arrangements.
- Items apart: total arrangements minus arrangements with them together, or place the others first and slot them into gaps.
- Arrangements in a circle: (n − 1)!.
Probability with combinations
P = favourable selections ÷ total selections.
Exam tip: Ask “does order matter?” first – it decides between permutations and combinations.
Check yourself
- In how many ways can 5 learners stand in a line?
- Evaluate ⁷C₃.
- In how many ways can 3 prefects be chosen from 10 learners?
Show answers
- 5! = 120
- 35
- ¹⁰C₃ = 120
Questions about this topic (0)
No questions about this topic yet.
Sign in or create a free account to join in.