Class 11 Maths Notes
~9 min readPermutations and Combinations Class 11 Notes
Complete, exam-ready notes on permutations and combinations: the fundamental principle of counting, factorials, arrangements (permutations) and selections (combinations), the relation between nPr and nCr, circular permutations, and arrangements of identical objects — written for CBSE boards and JEE revision.
What is the core idea of this chapter?
Permutations count ordered arrangements and combinations count unordered selections — both built on the fundamental principle of counting and factorials.
Fundamental Principle of Counting
Multiplication principle
If one task can be done in m ways and a second independent task can be done in n ways, the two tasks together can be done in m × n ways. This extends to any number of sequential tasks.
Addition principle
If one task can be done in m ways and a different, mutually exclusive task can be done in n ways, exactly one of the two tasks can be done in m + n ways.
Multiplication vs addition
Factorials and Permutations
Permutation
A permutation is an arrangement of r objects chosen from n distinct objects where order matters. The number of such arrangements is .
- Permutations with repetition allowed: each of the r positions has n choices, giving n^r arrangements.
- All n objects arranged in a row: n! permutations.
- Arranging n objects in a circle: (n − 1)! circular permutations (fixing one object eliminates rotational duplicates).
Permutations of Objects Not All Distinct
When n objects include p of one kind, q of another, r of a third, and so on, the number of distinct arrangements is . For example, the letters of MISSISSIPPI (M=1, I=4, S=4, P=2) give distinct arrangements.
Circular Permutations
Circular permutation
In a circular arrangement, rotations of the same order are considered identical. Fixing one object's position eliminates rotational symmetry, leaving (n − 1)! arrangements of the remaining n − 1 objects.
- If clockwise and anticlockwise order are both the same (e.g., a necklace), the count is (n − 1)!/2.
- If direction matters (e.g., people seated around a table with a distinguished side), the count is (n − 1)!.
- Fixing one object is the standard technique — it removes all rotational duplicates.
Combinations
Combination
A combination is a selection of r objects from n distinct objects where order does not matter. The number of such selections is .
- Relation between permutations and combinations:
- .
- Symmetry property:
- .
- Pascal's identity:
- .
- Sum of all combinations:
- .
Key Properties and Relations
- nPr = r! · nCr — permutations count order, combinations do not.
- nCr = nC(n − r) — choosing r to include equals choosing n − r to exclude.
- nCr + nC(r − 1) = (n + 1)Cr — Pascal's recurrence, the basis of Pascal's triangle.
- For r objects taken one at a time with repetition from n objects: n^r ways.
- When all n objects are taken (r = n): nPn = nCn = 1 arrangement/selection.
Order matters or not?
Solved Examples
Example: In how many ways can a committee of 4 men and 3 women be chosen from 7 men and 5 women?
Solution: Choose 4 men from 7: . Choose 3 women from 5: . By the multiplication principle, total = 35 × 10 = 350 ways.
Example: Find the number of permutations of the letters of the word ARRANGE.
Solution: ARRANGE has 7 letters with A repeated 2 times and R repeated 2 times. Distinct arrangements = .
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q1–Q31)
- 1) How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) repetition of the digits is allowed? (ii) repetition is not allowed?
- 2) How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
- 3) How many 4-letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
- 4) How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
- 5) A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
- 6) Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
- 7) Evaluate: (i) 8! (ii) 4! − 3!.
- 8) Is 3! + 4! = 7! ?
- 9) Compute 8! / (6! × 2!).
- 10) If 1/6! + 1/7! = x/8!, find x.
- 11) Evaluate n!/(n − r)! when (i) n = 6, r = 2; (ii) n = 9, r = 5.
- 12) How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
- 13) How many 4-digit numbers are there with no digit repeated?
- 14) How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
- 15) Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
- 16) From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
- 17) Find n if ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9.
- 18) Find r if (i) ⁵Pᵣ = 2 · ⁶Pᵣ₋₁, (ii) ⁵Pᵣ = ⁶Pᵣ₋₁.
- 19) How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
- 20) How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time, (ii) all letters are used at a time, (iii) all letters are used but the first letter is a vowel?
- 21) In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?
- 22) In how many ways can the letters of the word PERMUTATIONS be arranged if (i) words start with P and end with S, (ii) vowels are all together, (iii) there are always 4 letters between P and S?
- 23) If ⁿC₈ = ⁿC₂, find ⁿC₂.
- 24) Determine n if (i) ²ⁿC₃ : ⁿC₃ = 12 : 1, (ii) ²ⁿC₃ : ⁿC₃ = 11 : 1.
- 25) How many chords can be drawn through 21 points on a circle?
- 26) In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
- 27) Find the number of ways of selecting 9 balls from 6 red, 5 white and 5 blue balls if each selection consists of 3 balls of each colour.
- 28) Determine the number of 5-card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
- 29) In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl, if each cricket team of 11 must include exactly 4 bowlers?
- 30) A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
- 31) In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Answers at a glance
- 1Q1 → (i) 125, (ii) 60.
- 2Q2 → 108.
- 3Q3 → 5040.
- 4Q4 → 336.
- 5Q5 → 8.
- 6Q6 → 20.
- 7Q7 → (i) 40320, (ii) 18.
- 8Q8 → No (30 ≠ 5040).
- 9Q9 → 28.
- 10Q10 → x = 64.
- 11Q11 → (i) 30, (ii) 15120.
- 12Q12 → 504.
- 13Q13 → 4536.
- 14Q14 → 60.
- 15Q15 → 120 total, 48 even.
- 16Q16 → 56.
- 17Q17 → n = 9.
- 18Q18 → (i) r = 3, (ii) r = 4.
- 19Q19 → 40320.
- 20Q20 → (i) 360, (ii) 720, (iii) 240.
- 21Q21 → 33810.
- 22Q22 → (i) 1814400, (ii) 2419200, (iii) 25401600.
- 23Q23 → 45.
- 24Q24 → (i) n = 5, (ii) n = 6.
- 25Q25 → 210.
- 26Q26 → 40.
- 27Q27 → 2000.
- 28Q28 → 778320.
- 29Q29 → 3960.
- 30Q30 → 200.
- 31Q31 → 35.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1Write the value of 0!. (Ans: 1)
- 2State the formula for nPr. (Ans: n!/(n − r)!)
- 3Find C(4, 1). (Ans: 4)
- 4In how many ways can a committee of 2 be chosen from 5 people? (Ans: C(5, 2) = 10)
2 Mark Questions
- 1How many 3-digit numbers can be formed from the digits 1–5 without repetition? (Ans: 5 × 4 × 3 = 60)
- 2How many arrangements of the word 'APPLE' are there? (Ans: 5!/2! = 60)
- 3Find the number of diagonals of a hexagon using the combination formula. (Ans: C(6, 2) − 6 = 9)
3 Mark Questions
- 1If nP3 = 60, find n. (Ans: n!/(n−3)! = 60 ⇒ n = 5)
- 2A team of 3 boys and 2 girls is to be selected from 5 boys and 4 girls; in how many ways? (Ans: C(5,3) × C(4,2) = 10 × 6 = 60)
- 3How many 4-letter words can be formed from the letters of the word 'EXAMINATION' without repetition, taking vowels always together? (Ans: treat the 5 vowels {E,A,A,I,O} converted to distinct with repeats — count via cases)
4 Mark Questions
- 1In how many ways can the letters of the word 'INDEPENDENCE' be arranged (i) without restriction, (ii) when the vowels always come together, (iii) when the word starts with I and ends with E? (Ans: (i) 12!/(3!2!4!), (ii) vowel block count, (iii) fixed ends) Show the factorial divisions.
- 2Prove the identity C(n, r) = C(n, n − r) using the factorial definition, and verify with n = 10, r = 3. Then find the number of ways to form a hand of 3 cards from a 52-card pack. (Ans: C(52, 3))
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty with full solutions), then attempt the timed tests.
Covers all six topics: fundamental principle of counting; factorials and permutations; permutations of objects not all distinct; circular permutations; combinations; and key properties and relations.
Higher-difficulty set with multi-step problems (counting passwords, word arrangements, circular seating, committee selections, combinatorial identities) and full worked answers. Includes an answer key.
Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 4), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.
Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Factorial
Permutations
Combinations
nPr in terms of nCr
Symmetry of combinations
Pascal's identity
Identical objects
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- Distinguish between 'arrange' (permutation) and 'select' (combination) — this decides whether order matters.
- Circular permutation of n distinct objects = (n − 1)!; halve it if clockwise = anticlockwise.
- nC0 + nC1 + ... + nCn = 2^n — the total number of subsets of an n-element set.
- nCr = nC(n − r) lets you compute the smaller value when r > n/2.
- With identical objects, divide n! by the factorials of the repetition counts.
- Repetition allowed: n objects taken r at a time gives n^r arrangements.
- Pascal's identity nCr + nC(r−1) = (n+1)Cr is useful for quick computation and proofs.
FAQ
Frequently asked questions
What is the difference between permutations and combinations?
Permutations count arrangements where order matters (e.g. ABC ≠ BAC); combinations count selections where order does not matter (e.g. {A,B,C} = {B,A,C}).
How do you calculate circular permutations?
Fix one object to remove rotational symmetry, then arrange the remaining (n − 1) objects in a line: (n − 1)! arrangements. Halve it if clockwise and anticlockwise are considered the same.
When do we use permutations with repetition?
When each of the r positions can be filled independently by any of the n objects (e.g. forming a 3-digit number where digits can repeat), the count is n^r.
What is Pascal's identity?
nCr + nC(r − 1) = (n + 1)Cr. It connects adjacent entries in Pascal's triangle and is useful for simplifying sums and proving combinatorial identities.
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