Permutations and combinations problems: Class 11 practice
Use permutations ⁿPᵣ = n!/(n − r)! when ORDER matters (arrangements, ranks, codes) and combinations ⁿCᵣ = n!/[r!(n − r)!] when it does not (selections, teams, committees).. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Factorials
The idea behind permutations and combinations problems
Use permutations ⁿPᵣ = n!/(n − r)! when ORDER matters (arrangements, ranks, codes) and combinations ⁿCᵣ = n!/[r!(n − r)!] when it does not (selections, teams, committees).
Step-by-step method
- Ask: does order matter?
- Arrangement → ⁿPᵣ; selection → ⁿCᵣ.
- For 'and' multiply counts; for 'or' add them.
Worked example 1 (Basic)
In how many ways can a 1st, 2nd prize be given to 8 students (one prize each)?
- Order matters → 8P2
- = 8!/6! = 56
- Answer: 56
Worked example 2 (Standard)
How many ways can a team of 3 be chosen from 8 players?
- Order does not matter → 8C3
- = 8!/(3! × 5!) = 56
- Answer: 56
Worked example 3 (Challenge)
A committee of 2 boys and 2 girls is formed from 7 boys and 5 girls. How many committees are possible?
- Boys: 7C2 = 21
- Girls: 5C2 = 10
- Both together: 21 × 10 = 210
- Answer: 210
Common mistakes and misconceptions
- Using P for a team selection
- Double-counting by multiplying when cases overlap
Practice questions
- (Basic) In how many ways can a 1st, 2nd, 3rd and 4th prize be given to 11 students (one prize each)?
- (Basic) In how many ways can a 1st, 2nd prize be given to 9 students (one prize each)?
- (Standard) How many ways can a team of 2 be chosen from 10 players?
- (Standard) How many ways can a team of 3 be chosen from 12 players?
- (Challenge) A committee of 2 boys and 2 girls is formed from 7 boys and 3 girls. How many committees are possible?
- (Challenge) A committee of 2 boys and 2 girls is formed from 6 boys and 4 girls. How many committees are possible?
Answer key with full working
- Q1: Order matters → 11P4 → = 11!/7! = 7920 → Answer: 7920
- Q2: Order matters → 9P2 → = 9!/7! = 72 → Answer: 72
- Q3: Order does not matter → 10C2 → = 10!/(2! × 8!) = 45 → Answer: 45
- Q4: Order does not matter → 12C3 → = 12!/(3! × 9!) = 220 → Answer: 220
- Q5: Boys: 7C2 = 21 → Girls: 3C2 = 3 → Both together: 21 × 3 = 63 → Answer: 63
- Q6: Boys: 6C2 = 15 → Girls: 4C2 = 6 → Both together: 15 × 6 = 90 → Answer: 90
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 permutations and combinations problems 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.