Permutations and Combinations help students count arrangements and selections without listing every possibility. If order matters, it is a permutation. If order does not matter, it is a combination.
Permutation and combination class 11 questions look confusing at first because many problems sound similar. The trick is to ask one question before solving: does order matter?
If the problem is about arranging letters, forming numbers, seating people or assigning positions, use permutations. If the problem is about selecting a team, choosing cards or forming a committee, use combinations.
These Important Questions Class 11 Maths Chapter 6 are arranged from basic formulas to mixed application questions. The goal is to help students build speed, avoid formula confusion and practise board-style answers.
Key Takeaways from Class 11 Maths Permutations and Combinations
| Area |
What Students Should Revise |
| Chapter Name |
Permutations and Combinations |
| Chapter Number |
Chapter 6 |
| Main Idea |
Counting arrangements and selections |
| Important Formulas |
nPr, nCr, factorial notation |
| Core Concept |
Order matters in permutation, order does not matter in combination |
| Scoring Areas |
Number formation, word arrangements, committees, cards |
| Common Mistake |
Using nPr when nCr is needed |
| Best Practice |
Identify whether the question asks arrangement or selection |
Important Questions Class 11 Maths Chapter 6: Formula Sheet
Before solving permutation and combination questions class 11, students should revise these formulas.
| Concept |
Formula |
| Factorial |
n! = n × (n − 1) × (n − 2) × ... × 1 |
| Zero factorial |
0! = 1 |
| Permutation without repetition |
nPr = n! / (n − r)! |
| Permutation with repetition |
nʳ |
| Combination |
nCr = n! / r!(n − r)! |
| Relation between nPr and nCr |
nPr = nCr × r! |
| Repeated objects |
n! / p₁!p₂!...pₖ! |
| Complement property |
nCr = nC(n − r) |
NCERT defines permutation as an arrangement in a definite order and combination as selection where order is not important.
Important Questions Class 11 Maths Chapter-Wise
Permutation and Combination Class 11 Questions on Counting Principle
The fundamental principle of counting is the base of this chapter. If one task can happen in m ways and another task can happen in n ways after that, the total number of ways is m × n.
Q1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 if repetition is allowed?
Each place can be filled in 5 ways.
Hundreds place = 5 ways
Tens place = 5 ways
Units place = 5 ways
Total numbers = 5 × 5 × 5 = 125
Answer: 125 numbers can be formed.
Q2. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 if repetition is not allowed?
Hundreds place = 5 ways
Tens place = 4 ways
Units place = 3 ways
Total numbers = 5 × 4 × 3 = 60
Answer: 60 numbers can be formed.
Q3. How many 3-digit even numbers can be formed from 1, 2, 3, 4, 5, 6 if repetition is allowed?
The units place must be even, so it can be filled by 2, 4 or 6.
Units place = 3 ways
Hundreds place = 6 ways
Tens place = 6 ways
Total numbers = 6 × 6 × 3 = 108
Answer: 108 even numbers can be formed.
Q4. A coin is tossed 3 times. How many possible outcomes are there?
Each toss has 2 outcomes: head or tail.
Total outcomes = 2 × 2 × 2 = 8
Answer: There are 8 possible outcomes.
Permutations and Combinations Class 11 Questions on Factorials
Factorial questions are usually short and direct. Students should be quick with values such as 0!, 1!, 3!, 4!, 5! and 6!.
Q5. Evaluate 5!.
5! = 5 × 4 × 3 × 2 × 1 = 120
Answer: 5! = 120.
Q6. Evaluate 7! / 5!.
7! / 5! = 7 × 6 × 5! / 5!
= 7 × 6
= 42
Answer: 7! / 5! = 42.
Q7. Is 3! + 4! equal to 7!?
3! + 4! = 6 + 24 = 30
7! = 5040
Answer: No, 3! + 4! is not equal to 7!.
Q8. If nCr = nC2 and r = 8, find n.
Given nC8 = nC2.
Using nCr = nC(n − r), we get:
8 = n − 2
n = 10
Answer: n = 10.
Permutations and Combinations Important Questions on nPr
Permutation is used when order matters. Number formation, word arrangement, ranking, seating and assigning posts are common nPr questions.
Q9. Evaluate 6P3.
6P3 = 6! / (6 − 3)!
= 6! / 3!
= 6 × 5 × 4
= 120
Answer: 6P3 = 120.
Q10. How many 4-letter codes can be formed using the first 10 letters of the English alphabet if no letter is repeated?
A 4-letter code is an arrangement, so use permutation.
Number of codes = 10P4
= 10 × 9 × 8 × 7
= 5040
Answer: 5040 codes can be formed.
Q11. From 8 persons, in how many ways can a chairman and vice-chairman be chosen if one person cannot hold both posts?
The posts are different, so order matters.
Number of ways = 8P2
= 8 × 7
= 56
Answer: The chairman and vice-chairman can be chosen in 56 ways.
Q12. How many 4-digit numbers can be formed using digits 1 to 9 if no digit is repeated?
Number of ways = 9P4
= 9 × 8 × 7 × 6
= 3024
Answer: 3024 numbers can be formed.
Class 11 Permutation and Combination Extra Questions on Word Arrangements
Word arrangement questions are high-scoring but need careful reading. If letters repeat, divide by the factorials of repeated letters.
Q13. How many words can be formed using all letters of the word EQUATION, using each letter exactly once?
The word EQUATION has 8 distinct letters.
Number of arrangements = 8! = 40320
Answer: 40320 arrangements can be formed.
Q14. How many arrangements can be made from the letters of the word BOOK?
The word BOOK has 4 letters, with O repeated twice.
Number of arrangements = 4! / 2!
= 24 / 2
= 12
Answer: 12 arrangements can be made.
Q15. How many arrangements can be made from the letters of the word ALLAHABAD?
The word has 9 letters. A appears 4 times and L appears 2 times.
Number of arrangements = 9! / 4!2!
= 7560
Answer: 7560 arrangements can be made.
Q16. How many arrangements can be made from the letters of DAUGHTER if all vowels occur together?
The vowels are A, U and E. Treat them as one block.
Now there are 6 objects: one vowel block and 5 consonants.
These can be arranged in 6! ways. The vowels inside the block can be arranged in 3! ways.
Total arrangements = 6! × 3!
= 720 × 6
= 4320
Answer: 4320 arrangements can be made.
Important Questions of Permutation and Combination Class 11 on nCr
Combination is used when selection matters but order does not. Committee, team, card and group questions usually use nCr.
Q17. Evaluate 8C2.
8C2 = 8! / 2!6!
= 8 × 7 / 2
= 28
Answer: 8C2 = 28.
Q18. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Select 3 boys from 5 and 3 girls from 4.
Number of ways = 5C3 × 4C3
= 10 × 4
= 40
Answer: The team can be selected in 40 ways.
Q19. How many chords can be drawn through 21 points on a circle?
A chord is formed by choosing 2 points.
Number of chords = 21C2
= 21 × 20 / 2
= 210
Answer: 210 chords can be drawn.
Q20. A bag contains 5 black and 6 red balls. In how many ways can 2 black and 3 red balls be selected?
Number of ways = 5C2 × 6C3
= 10 × 20
= 200
Answer: The balls can be selected in 200 ways.
PNC Questions Class 11 on Cards and Committees
PNC questions class 11 often include cards, committees and selection under conditions. These questions test whether students can split cases correctly.
Q21. In how many ways can 4 cards be chosen from a pack of 52 cards?
Number of ways = 52C4
= 270725
Answer: 4 cards can be chosen in 270725 ways.
Q22. In how many ways can 4 cards of the same suit be chosen from a pack of 52 cards?
There are 4 suits. Each suit has 13 cards.
Number of ways = 4 × 13C4
= 2860
Answer: 4 cards of the same suit can be chosen in 2860 ways.
Q23. A committee of 3 persons is to be formed from 2 men and 3 women. In how many ways can it be done?
Total persons = 5
Number of committees = 5C3
= 10
Answer: The committee can be formed in 10 ways.
Q24. How many committees of 3 persons can be formed with 1 man and 2 women from 2 men and 3 women?
Number of ways = 2C1 × 3C2
= 2 × 3
= 6
Answer: The committee can be formed in 6 ways.
Permutation and Combination Class 11 Important Questions with Conditions
Condition-based questions are common in exams. Students should split the case before applying formulas.
Q25. How many words can be formed from MONDAY using 4 letters at a time if no letter is repeated?
MONDAY has 6 distinct letters.
Number of words = 6P4
= 6 × 5 × 4 × 3
= 360
Answer: 360 words can be formed.
Q26. How many words can be formed from MONDAY using all letters if the first letter is a vowel?
Vowels in MONDAY are O and A.
First place can be filled in 2 ways. Remaining 5 letters can be arranged in 5! ways.
Total words = 2 × 5!
= 2 × 120
= 240
Answer: 240 words can be formed.
Q27. How many 6-digit numbers can be formed from 0, 1, 3, 5, 7 and 9 if the number is divisible by 10 and no digit is repeated?
For divisibility by 10, the units digit must be 0.
Fix 0 at the units place. The remaining 5 digits can be arranged in 5! ways.
Total numbers = 5! = 120
Answer: 120 numbers can be formed.
Q28. In how many ways can 5 girls and 3 boys be seated in a row so that no two boys sit together?
Arrange 5 girls first: 5! ways.
This creates 6 gaps around the girls. Choose and arrange 3 boys in these gaps: 6P3 ways.
Total arrangements = 5! × 6P3
= 120 × 120
= 14400
Answer: They can be seated in 14400 ways.
Permutations and Combinations Class 11 Extra Questions for Practice
Use these permutations and combinations class 11 extra questions after revising the solved examples.
Q29. How many 5-digit telephone numbers can be formed using digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Q30. How many 3-digit even numbers can be formed using 1, 2, 3, 4, 6, 7 if no digit is repeated?
Q31. In how many ways can the letters of the word PERMUTATIONS be arranged if the words start with P and end with S?
Q32. How many selections of 5 cards from a deck of 52 cards contain exactly one king?
Q33. In how many ways can a student choose 5 courses if 9 courses are available and 2 specific courses are compulsory?
Q34. How many words can be formed using 2 vowels and 3 consonants from the word DAUGHTER?
Q35. In how many ways can 9 balls be selected from 6 red, 5 white and 5 blue balls if the selection has 3 balls of each colour?
Q36. How many arrangements of the word MISSISSIPPI are possible if all four I’s do not come together?
These question types match the NCERT exercise and miscellaneous exercise pattern from Chapter 6.
Common Mistakes in Permutation and Combination Questions Class 11
Students often know the formula but apply it in the wrong situation. This chapter rewards correct interpretation more than memorisation.
Permutation should be used when order changes the outcome. For example, AB and BA are different arrangements.
Combination should be used when order does not change the outcome. For example, selecting A and B is the same as selecting B and A.
Repeated letters must be handled carefully. For words like BOOK, ALLAHABAD or MISSISSIPPI, divide by the factorials of repeated letters.
In number formation, check whether 0 can be placed at the first position. If 0 comes first, the number may no longer have the required number of digits.
In “at least” or “at most” questions, split the problem into cases and add the answers.
Quick Revision Notes for Class 11 Maths Permutations and Combinations
Permutations and Combinations are counting techniques. They help count arrangements and selections without listing all possibilities.
The fundamental principle of counting is used when tasks happen one after another. Multiply the number of choices at each step.
Permutation means arrangement in a definite order. Combination means selection without considering order.
Use nPr for arrangements and nCr for selections. Use n! for arranging n distinct objects.
For repeated objects, divide by the factorials of repeated items. For example, if one letter appears 3 times, divide by 3!.