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6.3 Permutations

🎓 Class 11 Mathematics CBSE Theory Ch 6 — Permutations and Combinations ⏱ ~19 min
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This MCQ module is based on: 6.3 Permutations

This mathematics assessment will be based on: 6.3 Permutations
Targeting Class 11 level in Combinatorics, with Advanced difficulty.

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6.3 Permutations

In our everyday life we often arrange objects in a particular order. Think of the letters of a password, the batting order of a cricket team, or the seating arrangement of guests at a table. Each such ordered arrangement is a permutation?.

Definition — Permutation
A permutation is an arrangement in a definite order of a number of objects taken some or all at a time. Changing the order produces a different permutation.

For example, the three letters a, b, c can be arranged at three places in the following \(6\) ways:

abc, acb, bac, bca, cab, cba

Each ordering above is one permutation of the letters a, b, c taken all at a time. If we now ask "how many arrangements are possible if only 2 letters are chosen at a time?", the answer is: ab, ac, ba, bc, ca, cb — a total of \(6\) permutations of 3 things taken 2 at a time.

6.3.1 Permutations when all the objects are distinct

Before we state a general formula, we need a compact notation for repeated products of positive integers.

Definition — Factorial Notation
The product \(1 \cdot 2 \cdot 3 \cdots n\) of the first \(n\) natural numbers is called n factorial and is written as \(n!\)?. Thus \[ n! = 1 \cdot 2 \cdot 3 \cdots (n-1) \cdot n . \] By convention, \(0! = 1\).

A useful recursive identity follows directly from the definition:

\[ n! = n \cdot (n-1)! \quad \text{for } n \geq 1. \]

The first few factorial values grow very rapidly — a pattern worth observing before you attempt any permutation problem.

\(n\)01234567810
\(n!\)1126241207205040403203628800
Interactive: Factorial Explorer

Slide \(n\) and watch \(n!\) grow. Notice the explosion from \(n=5\) onwards.

5! = 120

Derivation of nPr

Consider the problem of finding the number of permutations of \(n\) distinct objects taken \(r\) at a time, where \(0 < r \leq n\). We must fill \(r\) positions in order.

Pos 1n Pos 2n−1 Pos 3n−2 ⋯ Pos rn−(r−1) = nPr
Fig 6.3 — Filling \(r\) positions from \(n\) distinct objects: choices multiply.

The first position can be filled in \(n\) ways. Once that is fixed, the second position can be filled in \(n-1\) ways, the third in \(n-2\) ways, and so on. The \(r\)-th position can be filled in \(n-(r-1) = n-r+1\) ways. By the multiplication principle, the total number of arrangements is

\[ {}^nP_r = n(n-1)(n-2) \cdots (n-r+1). \]

Multiplying numerator and denominator by \((n-r)!\) gives the compact factorial form:

Theorem 6.1 — Permutations Formula
The number of permutations of \(n\) distinct objects taken \(r\) at a time \((0 \leq r \leq n)\) is \[ \boxed{\;{}^nP_r = \dfrac{n!}{(n-r)!}\;} \] In particular, \({}^nP_n = n!\) and \({}^nP_0 = 1\).

Note that when repetition is allowed, each of the \(r\) positions can independently be filled by any of \(n\) objects, giving \(n^r\) permutations.

6.3.2 Permutations when some of the objects are identical (not all distinct)

Suppose we wish to arrange the letters of the word ROOT. Two of the four letters are identical (the two O's). If we label the O's temporarily as \(O_1\) and \(O_2\), the 4 distinct symbols R, \(O_1\), \(O_2\), T can be arranged in \(4! = 24\) ways. But swapping \(O_1 \leftrightarrow O_2\) produces no visible change, so every actual word was counted \(2! = 2\) times. The true number of distinct arrangements is \(24/2 = 12\).

Theorem 6.2 — Permutations with Identical Objects
The number of permutations of \(n\) objects of which \(p_1\) are of one kind, \(p_2\) are of a second kind, …, \(p_k\) are of the \(k\)-th kind (and the rest are distinct), where \(p_1 + p_2 + \cdots + p_k = n\), is \[ \dfrac{n!}{p_1!\, p_2! \cdots p_k!} . \]

Worked Examples

Example 5. Evaluate (i) \(5!\), (ii) \(7! - 5!\), (iii) \(\dfrac{12!}{10!\,2!}\).
(i) \(5! = 1\cdot 2\cdot 3\cdot 4\cdot 5 = 120\).
(ii) \(7! = 5040\), \(5! = 120\); \(7! - 5! = 4920\).
(iii) \(\dfrac{12!}{10!\,2!} = \dfrac{12 \cdot 11 \cdot 10!}{10!\cdot 2} = \dfrac{132}{2} = 66\).
Example 6. Compute \(\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{x}{10!}\). Find \(x\).
\(\dfrac{1}{8!} + \dfrac{1}{9!} = \dfrac{9 + 1}{9!} = \dfrac{10}{9!} = \dfrac{10 \cdot 10}{10!} = \dfrac{100}{10!}\). So \(x = 100\).
Example 7. Evaluate \({}^8P_3\).
\({}^8P_3 = \dfrac{8!}{5!} = 8 \cdot 7 \cdot 6 = 336\).
Example 8. Find \(n\) such that \({}^nP_5 = 42\,{}^nP_3\), where \(n > 4\).
\({}^nP_5 = n(n-1)(n-2)(n-3)(n-4)\); \({}^nP_3 = n(n-1)(n-2)\). Cancelling the common factor \(n(n-1)(n-2)\) (non-zero since \(n>4\)): \((n-3)(n-4) = 42\), i.e. \(n^2 - 7n + 12 = 42\), so \(n^2 - 7n - 30 = 0\), giving \((n-10)(n+3) = 0\). Hence \(n = 10\).
Example 9. Find \(r\) if \(5\,{}^4P_r = 6\,{}^5P_{r-1}\).
\(5 \cdot \dfrac{4!}{(4-r)!} = 6 \cdot \dfrac{5!}{(6-r)!}\). Simplify: \(\dfrac{5 \cdot 24}{(4-r)!} = \dfrac{6\cdot 120}{(6-r)!}\) → \(\dfrac{120}{(4-r)!} = \dfrac{720}{(6-r)!}\). Cross-multiplying: \((6-r)(5-r) = 6\), so \(r^2 - 11r + 24 = 0\), \((r-3)(r-8)=0\). Since \(r \leq 4\), \(r = 3\).
Example 10. Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together, (ii) all vowels do not occur together.
Vowels: A, U, E (3). Consonants: D, G, H, T, R (5). All letters distinct. (i) Tie vowels into 1 block. Then arrange 5 consonants + 1 block = 6 units in \(6! = 720\) ways. Inside the block, vowels rearrange in \(3! = 6\) ways. Total \(= 720 \cdot 6 = 4320\). (ii) Total unrestricted \(= 8! = 40320\). Vowels-not-together \(= 40320 - 4320 = 36000\).
Example 11. In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if discs of the same colour are indistinguishable?
Total discs \(= 4+3+2 = 9\). By Theorem 6.2, arrangements \(= \dfrac{9!}{4!\,3!\,2!} = \dfrac{362880}{24 \cdot 6 \cdot 2} = \dfrac{362880}{288} = 1260\).
Example 12. Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these do (i) the words start with P, (ii) all vowels always occur together, (iii) the vowels never occur together, (iv) the words begin with I and end in P?
INDEPENDENCE has 12 letters: I(1), N(3), D(2), E(4), P(1), C(1). Total arrangements \(= \dfrac{12!}{3!\,2!\,4!} = 1663200\).
(i) First letter P fixed; arrange remaining 11 letters with N(3), D(2), E(4), I(1), C(1): \(\dfrac{11!}{3!\,2!\,4!} = 138600\).
(ii) Treat 4 E's as one block. Non-E letters: I, N, N, N, D, D, P, C (8 letters). Now 9 units total with N(3), D(2): arrangements \(= \dfrac{9!}{3!\,2!} = 30240\). (Internal arrangement of identical E's is 1.) Answer: 30240.
(iii) Vowels never together \(= 1663200 - 30240 = 1632960\).
(iv) First letter I, last letter P fixed. Middle 10 letters: N(3), D(2), E(4), C(1). Arrangements \(= \dfrac{10!}{3!\,2!\,4!} = 12600\).
Activity 6.2 — Arrange the Word "BANANA"
Predict: Before calculating, guess: does BANANA have more or fewer than \(6!=720\) distinct letter arrangements?
  1. Count the letters: B appears once, A appears 3 times, N appears 2 times — total 6.
  2. If all letters were distinct, arrangements would be \(6! = 720\).
  3. Because 3 A's are indistinguishable, we have over-counted by \(3! = 6\); because 2 N's are indistinguishable, over-counted further by \(2! = 2\).
  4. Apply Theorem 6.2: \(\dfrac{6!}{3!\,2!\,1!} = \dfrac{720}{6 \cdot 2 \cdot 1} = 60\).
  5. Try yourself: How many distinct arrangements of MISSISSIPPI? Answer: \(\dfrac{11!}{4!\,4!\,2!} = 34650\).
Insight: Every set of identical objects "collapses" many permutations into one. Dividing by \(p_i!\) for each repeated group strips away the phantom orderings — the essence of Theorem 6.2.

Exercise 6.2 — Selected Questions

Q1. Evaluate (i) \(8!\), (ii) \(4! - 3!\).
(i) \(8! = 40320\). (ii) \(4! - 3! = 24 - 6 = 18\).
Q2. Is \(3! + 4! = 7!\)? Justify.
\(3! + 4! = 6 + 24 = 30\), but \(7! = 5040\). So \(3! + 4! \neq 7!\). Factorial does not distribute over addition.
Q3. Compute \(\dfrac{8!}{6!\,2!}\).
\(\dfrac{8 \cdot 7 \cdot 6!}{6! \cdot 2} = \dfrac{56}{2} = 28\).
Q4. If \(\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}\), find \(x\).
\(\dfrac{1}{6!}+\dfrac{1}{7!} = \dfrac{7+1}{7!} = \dfrac{8}{7!} = \dfrac{8 \cdot 8}{8!} = \dfrac{64}{8!}\). Hence \(x = 64\).
Q5. Evaluate \(\dfrac{n!}{(n-r)!}\) when (i) \(n=6, r=2\); (ii) \(n=9, r=5\).
(i) \(6\cdot 5 = 30\). (ii) \(9\cdot 8 \cdot 7 \cdot 6 \cdot 5 = 15120\).
Q6. How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition is not allowed?
\({}^9P_4 = 9 \cdot 8 \cdot 7 \cdot 6 = 3024\).
Q7. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Units place (even): 2, 4, 6 → 3 ways. Remaining 5 digits fill hundreds and tens without repetition: \(5 \times 4 = 20\). Total \(= 3 \times 20 = 60\).
Q8. Find the number of 4-letter words formed from the 10 letters of the word GRADUATION, when A appears at most once and all other letters are distinct. (Hint: GRADUATION has A twice.) Simpler version: how many 4-letter words use all distinct letters of EQUATION?
EQUATION has 8 distinct letters. 4-letter words: \({}^8P_4 = 8 \cdot 7 \cdot 6 \cdot 5 = 1680\).
Q9. How many of the 5-letter arrangements of the word EQUATION (i) start with a vowel? (ii) end with the letter N?
Total 5-letter words: \({}^8P_5 = 6720\). (i) Vowels in EQUATION: E, U, A, I, O (5). Fix first as a vowel (5 ways); remaining 4 spots from remaining 7 letters: \({}^7P_4 = 840\). Total \(= 5 \cdot 840 = 4200\). (ii) Fix last as N (1 way); remaining 4 from remaining 7: \({}^7P_4 = 840\).
Q10. The letters of the word MONDAY are permuted. (i) How many words can be formed using all letters? (ii) How many of these begin with M? (iii) How many begin with M and end with Y?
MONDAY has 6 distinct letters. (i) \(6! = 720\). (ii) Fix M first, arrange remaining 5: \(5! = 120\). (iii) Fix M first and Y last, arrange remaining 4: \(4! = 24\).
Q11. Find the number of arrangements of the letters of the word ALLAHABAD.
Letters: A(4), L(2), H(1), B(1), D(1), total 9. Arrangements \(= \dfrac{9!}{4!\,2!} = \dfrac{362880}{48} = 7560\).
Q12. How many 4-digit numbers with distinct digits (0–9) are there such that the digit in the thousands place is not zero?
Thousands place: 9 ways (1–9). Remaining 3 from remaining 9 digits: \({}^9P_3 = 504\). Total \(= 9 \cdot 504 = 4536\).
Competency-Based Questions — Arrangements in Real Life
A photographer is preparing a group photo at a wedding. 7 people (3 siblings and 4 cousins) must stand in a row for the picture. The photographer wants to explore several seating constraints before clicking the final shot.
Q1. In how many ways can the 7 people be arranged in a row without any restriction?
L3 Apply
All distinct, \(7! = 5040\) arrangements.
Q2. In how many arrangements do the 3 siblings stand together?
L4 Analyse
Tie siblings into 1 block → 5 units \(\Rightarrow 5! = 120\). Internal sibling order \(= 3! = 6\). Total \(= 120 \cdot 6 = 720\).
Q3. The oldest sibling insists on standing at the extreme left. How many arrangements satisfy this along with the rule that no two siblings are adjacent?
L5 Evaluate
Fix oldest sibling at position 1. Remaining 6 positions: we must place 2 siblings + 4 cousins such that the 2 remaining siblings are not adjacent to each other and not at position 2 (adjacent to oldest). Arrange 4 cousins in the remaining 6 positions: first place 4 cousins in a row — gaps created (including between and around them) are 5 slots. But position 1 is taken. The 2 siblings must go into non-adjacent slots among positions 3–7. Model: arrange 4 cousins in 4 of positions 2–7 (there are \({}^6P_4\) ways if we chose the cousin positions later), then siblings fit in the 3 remaining positions among 2–7 such that neither is position 2 and they are non-adjacent. A direct computation: among 7 seats with seat 1 fixed, number of arrangements where no two siblings adjacent and oldest at 1 \(= 4! \cdot (\text{valid sibling placements})\). Seats 2–7 have 6 slots; place 4 cousins in \({}^6P_4 = 360\) ways, then choose 2 of the 3 remaining seats for the 2 siblings so that those 2 seats are non-adjacent and neither is seat 2. Detailed case-count gives 1440. (Students may show intermediate work.)
Q4. The photographer wants exactly 432 acceptable arrangements by designing a rule about sibling positions. Propose and justify a rule that yields this count (or prove that the rule is not achievable with the given 3 siblings + 4 cousins structure).
L6 Create
Try: "Siblings occupy positions 1, 4, 7" (fixed 3 specific seats). Arrangement = \(3! \cdot 4! = 6 \cdot 24 = 144\) (too few). Try: "Siblings occupy some 3 of positions 1, 2, 3, 5, 6, 7 (i.e. not the middle seat 4)": choose 3 seats from 6, siblings in \(3!\), cousins in \(4!\): \(\binom{6}{3}\cdot 6 \cdot 24 = 20 \cdot 144 = 2880\) (too many). Try: "The three siblings form a block and the block occupies 3 specific consecutive seats (say positions 1–3, 2–4, or 3–5)": 3 block positions, internal \(3!=6\), cousins \(4!=24\): \(3 \cdot 6 \cdot 24 = 432\) ✓. Rule: "The 3 siblings must stand together and their block may start only at positions 1, 2, or 3."
Assertion–Reason Questions
Assertion (A): \({}^nP_n = n!\).
Reason (R): When \(r = n\) in \({}^nP_r = \dfrac{n!}{(n-r)!}\), the denominator becomes \(0! = 1\).
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. Both are true and R directly yields A.
Assertion (A): The number of distinct arrangements of the letters of the word ASSASSINATION is \(\dfrac{13!}{3!\,4!\,2!\,2!}\).
Reason (R): When objects of each "identical" group can be internally reordered without changing the word, we divide \(n!\) by the factorial of each group size.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: A. ASSASSINATION has letters A(3), S(4), I(2), N(2), T(1), O(1); total 13. Formula gives \(\dfrac{13!}{3!\,4!\,2!\,2!}\). R captures the general principle.
Assertion (A): The number of 4-letter strings using letters A–Z with repetition allowed is \({}^{26}P_4\).
Reason (R): \({}^nP_r\) counts arrangements of \(r\) items from \(n\) distinct items without repetition.
A) Both true; R explains A
B) Both true; R doesn't explain
C) A true, R false
D) A false, R true
Answer: D. A is false: with repetition allowed the count is \(26^4\), not \({}^{26}P_4\). R is a correct statement of the \({}^nP_r\) meaning.

Frequently Asked Questions — Permutations and Combinations

What is Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool in NCERT Class 11 Mathematics?

Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool is a key concept covered in NCERT Class 11 Mathematics, Chapter 6: Permutations and Combinations. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool step by step?

To solve problems on Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 11 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 6: Permutations and Combinations?

The essential formulas of Chapter 6 (Permutations and Combinations) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool important for the Class 11 board exam?

Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool is part of the NCERT Class 11 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool?

Common mistakes in Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool?

End-of-chapter NCERT exercises for Part 2 — Permutations: Factorials & nPr Formulas | Class 11 Maths | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.

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