Arrangements with repeated letters: Class 11 practice
The number of arrangements of n objects where some repeat is n! divided by the factorial of each repeat count. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Factorials
The idea behind arrangements with repeated letters
The number of arrangements of n objects where some repeat is n! divided by the factorial of each repeat count. Conditions like 'vowels together' are handled by treating a group as one block.
Step-by-step method
- Count total letters and each repeat.
- Divide n! by the repeats' factorials.
- For 'together' conditions, treat the group as one unit and multiply by its internal arrangements.
Worked example 1 (Basic)
How many different arrangements of the letters of ARRANGE are there?
- 7 letters; repeats: A×2, R×2
- 7!/4 = 1260
- Answer: 1260
Worked example 2 (Standard)
How many different arrangements of the letters of STATISTICS are there?
- 10 letters; repeats: S×3, T×3, I×2
- 10!/72 = 50400
- Answer: 50400
Worked example 3 (Challenge)
How many different arrangements of the letters of PEPPER are there?
- 6 letters; repeats: P×3, E×2
- 6!/12 = 60
- Answer: 60
Common mistakes and misconceptions
- Forgetting to divide for repeats
- Not arranging letters inside a block
Practice questions
- (Basic) How many different arrangements of the letters of COMMITTEE are there?
- (Basic) How many different arrangements of the letters of APPLE are there?
- (Standard) How many different arrangements of the letters of TOFFEE are there?
- (Standard) How many different arrangements of the letters of MISSISSIPPI are there?
- (Challenge) How many different arrangements of the letters of LETTER are there?
- (Challenge) How many different arrangements of the letters of BALLOON are there?
Answer key with full working
- Q1: 9 letters; repeats: M×2, T×2, E×2 → 9!/8 = 45360 → Answer: 45360
- Q2: 5 letters; repeats: P×2 → 5!/2 = 60 → Answer: 60
- Q3: 6 letters; repeats: F×2, E×2 → 6!/4 = 180 → Answer: 180
- Q4: 11 letters; repeats: I×4, S×4, P×2 → 11!/1152 = 34650 → Answer: 34650
- Q5: 6 letters; repeats: E×2, T×2 → 6!/4 = 180 → Answer: 180
- Q6: 7 letters; repeats: L×2, O×2 → 7!/4 = 1260 → Answer: 1260
Syllabus note
This skill is listed under "Permutations and Combinations" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is arrangements with repeated letters in?
"Permutations and Combinations", Class 11 Maths (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.