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Permutations and Combinations

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Chapter Notes

Overview
This chapter introduces the fundamental principles of counting, which are essential for determining the number of ways events can occur without listing all possibilities. It covers the multiplication principle, permutations (arrangements where order matters), and combinations (selections where order does not matter). The one big idea a student must walk away with is that permutations are for ordered arrangements and combinations are for unordered selections, with the key formula linking them being nPr = nCr × r!.
Key Concepts (10)
Fundamental Principle of Counting (Multiplication Principle)
If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrences of the events in the given order is m × n.
e.g. Mohan has 3 pants and 2 shirts. The number of different pairs of a pant and a shirt he can dress up with is 3 × 2 = 6.
Permutation
A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.
e.g. The number of 3-letter words that can be formed from the letters of the word NUMBER without repetition is 6 × 5 × 4 = 120.
Factorial notation (n!)
The product of the first n natural numbers, i.e., 1 × 2 × 3 × ... × (n-1) × n, is denoted as n! and read as 'n factorial'. 0! is defined as 1.
e.g. 5! = 1 × 2 × 3 × 4 × 5 = 120.
nPr (Permutations of n different objects taken r at a time)
The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat, is n(n-1)(n-2)...(n-r+1), which is denoted by nPr.
e.g. The number of ways to choose a Chairman and Vice-Chairman from 12 persons is 12P2 = 12 × 11 = 132.
Combination
A combination is a selection of objects where the order is not important.
e.g. From a group of 3 lawn tennis players X, Y, Z, a team of 2 players can be formed in 3 ways: XY, YZ, ZX.
nCr (Combinations of n different objects taken r at a time)
The number of combinations of n different objects taken r at a time, where 0 ≤ r ≤ n, is given by nCr = n! / [r!(n-r)!].
e.g. The number of ways to select a committee of 3 persons from a group of 5 is 5C3 = 10.
Permutations when all objects are not distinct
The number of permutations of n objects, where p1 objects are of one kind, p2 are of second kind, ..., pk are of kth kind, and the rest, if any, are of different kind, is n! / (p1! p2! ... pk!).
e.g. The number of distinct permutations of the letters of the word ALLAHABAD (9 letters: 4 A's, 2 L's) is 9! / (4! 2!) = 7560.
Permutations with repetition allowed
The number of permutations of n different objects taken r at a time, where repetition is allowed, is n^r.
e.g. The number of 4-letter words that can be formed from the letters of the word ROSE with repetition allowed is 4^4 = 256.
nP0
The number of permutations of n different objects taken 0 at a time is defined as 1, i.e., nP0 = 1.
e.g. Arranging no object at all is the same as leaving behind all objects, and there is only one way of doing so.
nC0
The number of combinations of n different things taken nothing at all is defined as 1, i.e., nC0 = 1.
e.g. Selecting nothing at all is the same as leaving behind all objects, and there is only one way of doing so.
Formulas & Equations (10)
nPr = n(n-1)(n-2)...(n-r+1)
Reads as: n P r equals n times n minus 1 times n minus 2, and so on, up to n minus r plus 1
Variables: n = total number of distinct objects, r = number of objects taken at a time (0 < r ≤ n)
Use when: Use when counting the number of arrangements (ordered selections) of r distinct objects from n distinct objects without repetition.
⚠ Students often forget to stop at (n-r+1) and instead write n(n-1)(n-2)...(n-r), which is incorrect.
nPr = n! / (n-r)!
Reads as: n P r equals n factorial divided by n minus r factorial
Variables: n = total number of distinct objects, r = number of objects taken at a time (0 ≤ r ≤ n)
Use when: Use this factorial form of the permutation formula for easier calculation, especially when n and r are large.
⚠ Students often incorrectly write nPr = n! / r! instead of n! / (n-r)!.
n! = n × (n-1)!
Reads as: n factorial equals n times n minus 1 factorial
Variables: n = a natural number (n ≥ 1)
Use when: Use this recursive property to simplify factorial expressions, e.g., 7! = 7 × 6!.
⚠ Students sometimes think n! = n × (n-1), forgetting the factorial on (n-1).
0! = 1
Reads as: zero factorial equals one
Variables: No variables; it is a definition.
Use when: Use this definition when applying the formula nPr = n!/(n-r)! for r = n, or in any combinatorial expression where 0! appears.
⚠ Students often think 0! = 0, which leads to division by zero errors.
Number of permutations of n objects taken r at a time with repetition allowed = n^r
Reads as: n raised to the power r
Variables: n = number of distinct objects, r = number of selections/positions
Use when: Use when each of the r positions can be filled by any of the n objects, and repetition is allowed.
⚠ Students confuse this with nPr and use the permutation formula even when repetition is allowed.
nCr = n! / [r!(n-r)!]
Reads as: n C r equals n factorial divided by r factorial times n minus r factorial
Variables: n = total number of distinct objects, r = number of objects selected (0 ≤ r ≤ n)
Use when: Use when counting the number of ways to select r objects from n distinct objects where order does not matter.
⚠ Students often forget the r! in the denominator, writing nCr = n!/(n-r)! (which is actually nPr).
nCr = nC(n-r)
Reads as: n C r equals n C n minus r
Variables: n = total number of distinct objects, r = number of objects selected
Use when: Use to simplify calculations, e.g., 50C48 = 50C2, which is easier to compute.
⚠ Students incorrectly think nCr = nC(n-r) implies r = n-r always, forgetting the property is about equality of values, not indices.
nCr + nC(r-1) = (n+1)Cr
Reads as: n C r plus n C r minus 1 equals n plus 1 C r
Variables: n = total number of distinct objects, r = number of objects selected
Use when: Use this Pascal's rule to simplify sums of combinations or to derive combinatorial identities.
⚠ Students often misapply the indices, e.g., writing nCr + nC(r+1) = (n+1)C(r+1) incorrectly.
nPr = nCr × r!
Reads as: n P r equals n C r times r factorial
Variables: n = total number of distinct objects, r = number of objects taken
Use when: Use to relate permutations and combinations, or to derive one from the other.
⚠ Students often forget the r! factor and think nPr = nCr.
Number of permutations of n objects where p1 are of one kind, p2 of second kind, ..., pk of kth kind = n! / (p1! p2! ... pk!)
Reads as: n factorial divided by p1 factorial times p2 factorial, and so on, up to pk factorial
Variables: n = total number of objects, p1, p2, ..., pk = numbers of objects of each identical kind
Use when: Use when arranging objects where some are identical, e.g., rearranging letters of a word with repeated letters.
⚠ Students forget to divide by the factorial of the count of each repeated object, leading to overcounting.
How It Works
Applying the Multiplication Principle
  1. Step 1: Identify the sequence of events or choices to be made.
  2. Step 2: Determine the number of ways each event can occur.
  3. Step 3: Multiply the number of ways for each event to get the total number of outcomes.
💡 The multiplication principle applies only when events occur in a specific order, and each choice is independent of the previous ones.
Counting Permutations (nPr) using the formula
  1. Step 1: Identify n (total distinct objects) and r (number to arrange).
  2. Step 2: Use the formula nPr = n! / (n-r)! or the product form n(n-1)...(n-r+1).
  3. Step 3: Compute the value, ensuring 0 ≤ r ≤ n.
💡 The order of arrangement matters in permutations. For r = n, nPn = n!.
Counting Combinations (nCr) using the formula
  1. Step 1: Identify n (total distinct objects) and r (number to select).
  2. Step 2: Use the formula nCr = n! / [r!(n-r)!].
  3. Step 3: Compute the value, ensuring 0 ≤ r ≤ n.
💡 Order does not matter in combinations. Selecting r objects is the same as rejecting (n-r) objects, so nCr = nC(n-r).
Counting Permutations of Objects with Repetition (identical objects)
  1. Step 1: Count the total number of objects (n).
  2. Step 2: Identify the counts of each type of identical object (p1, p2, ..., pk).
  3. Step 3: Apply the formula: number of arrangements = n! / (p1! p2! ... pk!).
💡 This formula is used only when objects are not all distinct. Do not use it when all objects are different.
Using the Addition Principle (when events are mutually exclusive)
  1. Step 1: Identify mutually exclusive cases (e.g., signals with 2 flags, 3 flags, etc.).
  2. Step 2: Count the number of ways for each case separately.
  3. Step 3: Add the numbers from all cases to get the total number of ways.
💡 The addition principle is used when events cannot happen simultaneously. In Example 4, signals of different lengths are counted separately and then added.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Permutation and Combination
✗ Students write nPr when the question asks for selection (order not important), e.g., using 5P3 instead of 5C3 for selecting a committee.
→ Permutations count ordered arrangements, while combinations count unordered selections. A committee of 3 people is a selection, not an arrangement.
✓ Identify if order matters. If it does (e.g., arranging letters, assigning positions), use nPr. If not (e.g., selecting a team, choosing cards), use nCr.
⚠ Forgetting to Divide by Factorials for Identical Objects
✗ For the word 'ALLAHABAD', students write 9! as the number of arrangements, ignoring the repeated A's and L's.
→ Treating identical letters as distinct overcounts arrangements. Swapping two identical A's does not create a new word.
✓ Use the formula for permutations of objects with repetition: number of arrangements = 9! / (4! 2!) = 7560.
⚠ Misapplying the Multiplication Principle with Constraints
✗ For forming 2-digit even numbers from digits 1,2,3,4,5 with repetition, students fill the tens place first (5 ways) and then the units place (2 ways), getting 5×2=10, but they sometimes write 2×5=10 without justification.
→ The order of filling places matters only if constraints affect the count. Here, the units place has a constraint (must be even), so it should be filled first. The answer is correct, but the reasoning must show awareness of the constraint.
✓ Since the number must be even, the units place can only be filled by 2 or 4 (2 ways). Then the tens place can be filled by any of the 5 digits (5 ways). So total = 2 × 5 = 10.
⚠ Incorrectly Handling 'At Least' or 'At Most' Conditions
✗ For selecting a team of 5 from 4 girls and 7 boys with at least 1 girl, students write 4C1 × 10C4, which overcounts.
→ This approach selects 1 girl first and then selects the remaining 4 from the remaining 10 people, but it counts the same team multiple times (e.g., selecting Girl A first and then Girl B later is the same as selecting Girl B first and then Girl A).
✓ Break into mutually exclusive cases: (1 girl, 4 boys) + (2 girls, 3 boys) + (3 girls, 2 boys) + (4 girls, 1 boy). Use combinations for each case and add.
⚠ Forgetting that 0! = 1
✗ When computing nPn = n! / 0!, students write n! / 0, leading to an undefined result.
→ 0! is defined as 1, not 0. The formula nPn = n! / 0! = n! / 1 = n! is correct.
✓ Memorize that 0! = 1. Always use this value when simplifying expressions involving 0!.
⚠ Using nCr = n!/(n-r)! Instead of n!/(r!(n-r)!)
✗ Students write nCr = n! / (n-r)!, which is actually the formula for nPr.
→ The formula for combinations must include r! in the denominator to account for the fact that order does not matter.
✓ Write the correct formula: nCr = n! / [r! (n-r)!]. Remember: 'r!' in the denominator for combinations.
Compare & Contrast
Order of objects
Permutation (nPr)
Combination (nCr)
In permutations, order matters (e.g., arranging letters to form words). In combinations, order does not matter (e.g., selecting a committee).
Whether an object can be used more than once
Permutations with repetition allowed
Permutations without repetition (nPr)
With repetition allowed, the number of permutations is n^r (each of the r positions has n choices). Without repetition, the number is nPr = n!/(n-r)! (choices decrease by 1 each time).
Exam Focus
Marks pattern: 1-2 MCQs (1 mark each), 1-2 Very Short Answer (1-2 marks), 1 Short Answer (3 marks), 1 Long Answer (5 marks). Total approximately 8-12 marks.
Most asked: Finding the number of arrangements of letters of a word with repeated letters · Selecting teams or committees with constraints (e.g., at least one girl, exactly 4 bowlers) · Forming numbers with given digits (with or without repetition, with conditions like even numbers) · Using nCr = nC(n-r) to simplify calculations · Problems involving both permutations and combinations (e.g., selecting and then arranging)
Don't skip: Long answer questions (5 marks) involving multi-step selection and arrangement, e.g., 'How many words can be formed from the letters of a given word with conditions on vowels/consonants?'; Problems on permutations of objects with repetition (e.g., rearranging letters of words like INDEPENDENCE, MISSISSIPPI)
One-Line Revision (read aloud before the exam)
  1. The Fundamental Principle of Counting states: if one event can occur in m ways and another in n ways, the total number of ways both occur in order is m × n.
  2. A permutation is an arrangement where order matters; a combination is a selection where order does not matter.
  3. n! (n factorial) is the product of the first n natural numbers, and 0! is defined as 1.
  4. The number of permutations of n distinct objects taken r at a time is nPr = n!/(n-r)!.
  5. The number of combinations of n distinct objects taken r at a time is nCr = n!/(r!(n-r)!).
  6. nCr = nC(n-r): selecting r objects is the same as rejecting (n-r) objects.
  7. nPr = nCr × r!: every combination of r objects can be arranged in r! ways to give permutations.
  8. The number of permutations of n objects with p1 identical of one kind, p2 of another, etc., is n!/(p1! p2! ... pk!).
  9. If repetition is allowed, the number of permutations of n objects taken r at a time is n^r.
  10. When solving 'at least' or 'at most' problems, break into mutually exclusive cases and add the combinations.
  11. Pascal's rule: nCr + nC(r-1) = (n+1)Cr.
  12. For forming numbers with digits, always consider constraints like 'even' or 'greater than' before filling places.
  13. In problems with both selection and arrangement, first select (using combinations), then arrange (using permutations).
  14. nC0 = 1 and nP0 = 1: there is exactly one way to select or arrange nothing.

MCQs — 10 Questions with Answers

Q1. How many 3-digit numbers can be formed from digits 1, 2, 3, 4, 5 if repetition of digits is not allowed?

  • A. 60 ✓
  • B. 125
  • C. 120
  • D. 20

Answer: A — Using multiplication principle: first digit (5 choices) × second digit (4 remaining) × third digit (3 remaining) = 5 × 4 × 3 = 60.

Q2. How many 3-digit numbers can be formed from digits 1, 2, 3, 4, 5 if repetition of digits is allowed?

  • A. 60
  • B. 125 ✓
  • C. 120
  • D. 25

Answer: B — With repetition: each of 3 positions independently has 5 choices, so 5 × 5 × 5 = 125.

Q3. How many 3-digit even numbers can be formed from digits 1, 2, 3, 4, 5, 6 if digits can be repeated?

  • A. 108 ✓
  • B. 90
  • C. 180
  • D. 36

Answer: A — Fill units place first (even: only 2, 4, 6 → 3 choices) × hundreds place (any 6 digits) × tens place (any 6 digits) = 3 × 6 × 6 = 108.

Q4. From 4 different coloured flags, how many 2-flag signals can be generated if signals are distinguished by order (one flag above the other)?

  • A. 6
  • B. 8
  • C. 12 ✓
  • D. 16

Answer: C — Upper position (4 choices) × lower position (3 remaining) = 4 × 3 = 12 because order matters and no repetition.

Q5. How many different ways can Mohan choose one pant and one shirt if he has 3 pants and 2 shirts?

  • A. 5
  • B. 6 ✓
  • C. 3
  • D. 2

Answer: B — Pant (3 choices) × shirt (2 choices) = 3 × 2 = 6 by multiplication principle.

Q6. A student must fill 4 vacant places of a code using first 10 English letters with no letter repeated. How many codes are possible?

  • A. 10000
  • B. 5040 ✓
  • C. 4020
  • D. 40

Answer: B — First place (10 choices) × second (9) × third (8) × fourth (7) = 10 × 9 × 8 × 7 = 5040.

Q7. Which statement is NOT correct regarding the Fundamental Principle of Counting?

  • A. It applies when events occur in succession
  • B. For n mutually exclusive events, we add their counts instead of multiplying
  • C. It requires all choices to remain constant across all positions ✓
  • D. It allows us to count without listing all possibilities

Answer: C — Statement C is false because choices CAN change (and usually do): with repetition allowed choices stay same, but without repetition, choices decrease at each position.

Q8. Two events A and B are such that A can occur in 5 ways and B can occur in 3 ways. If neither event restricts the other, then the total number of ways both can occur together is ___. (Both/Neither/One is true: In this case both the multiplication principle and outcome counting are equivalent)

  • A. 5 + 3 = 8
  • B. 5 × 3 = 15 ✓
  • C. 5 − 3 = 2
  • D. 5² + 3² = 34

Answer: B — Since events are independent and both occur (not mutually exclusive), multiplication principle applies: 5 × 3 = 15 total combined outcomes.

Q9. How many signals can be generated using at least 2 flags from 5 different flags available (order matters, one flag below the other)?

  • A. 80
  • B. 200
  • C. 300
  • D. 320 ✓

Answer: D — Add all cases: (2-flag: 5×4=20) + (3-flag: 5×4×3=60) + (4-flag: 5×4×3×2=120) + (5-flag: 5×4×3×2×1=120) = 320.

Q10. A lock has 4 wheels labelled with digits 0–9. If the first digit is fixed as 7 and no digit repeats, how many possible sequences must be checked for the remaining 3 wheels?

  • A. 720
  • B. 504 ✓
  • C. 1000
  • D. 243

Answer: B — First digit fixed as 7. Remaining 3 wheels choose from 9 remaining digits (0-9 excluding 7) without repetition: 9 × 8 × 7 = 504.

Flashcards

State the Fundamental Principle of Counting for two successive events.

If event 1 can occur in m ways and event 2 in n ways, total occurrences in order = m × n.

How many 4-letter words (no repetition) from letters of ROSE?

4 × 3 × 2 × 1 = 24 because each position has progressively fewer available letters.

How many 4-letter words (repetition allowed) from letters of ROSE?

4 × 4 × 4 × 4 = 256 because each position independently has all 4 letters available.

What is the key difference between filling tens place vs units place in 2-digit even numbers?

Units place must be filled first (restricted to 2 or 4) to minimise wasted cases; tens place is filled second with all available digits.

How many 2-flag signals from 5 different flags (order matters)?

5 × 4 = 20 because first flag has 5 choices and second has 4 remaining choices.

For 'at least 2 flags' problem with 5 flags, why do we add separate case counts?

Because '2 flags' and '3 flags' are mutually exclusive events; total = (2-flag count) + (3-flag count) + (4-flag count) + (5-flag count).

How many 4-letter codes from first 10 English letters (no repetition)?

10 × 9 × 8 × 7 = 5040 because each position loses one available letter.

In Sabnam's problem (2 bags, 3 boxes, 2 bottles), why multiply all numbers?

Every choice of bag (2 ways) × every choice of box (3 ways) × every choice of bottle (2 ways) = 2 × 3 × 2 = 12 total combinations.

What changes when digits 'can be repeated' vs 'cannot be repeated' in forming numbers?

Without repetition: each digit fills a position then becomes unavailable, so choices decrease; with repetition: each position has same number of choices throughout.

How many 5-flag signals (order matters) from 5 different flags?

5 × 4 × 3 × 2 × 1 = 120 because all flags are used in descending choice sequence.

Important Board Questions

State the Fundamental Principle of Counting and illustrate it with the example of Mohan choosing one pant from 3 and one shirt from 2. [2 marks]

Define the principle clearly: if event 1 occurs in m ways and event 2 in n ways, total = m × n. Then apply: 3 pants × 2 shirts = 6 combinations and list them (P₁S₁, P₁S₂, P₂S₁, P₂S₂, P₃S₁, P₃S₂).

Find the number of 4-letter words (with or without meaning) that can be formed from the letters of ROSE if repetition of letters is not allowed. Show all working steps and explain why each step reduces the number of choices. [5 marks]

Apply multiplication principle to 4 consecutive positions: position 1 (4 letters available) × position 2 (3 remaining, because 1 used) × position 3 (2 remaining) × position 4 (1 remaining). Explain that 'no repetition' forces each subsequent position to have one fewer choice than the previous, resulting in 4! = 24.

From 5 different flags, how many signals can be generated if at least 2 flags are used in order (one below the other)? Solve by breaking into cases, showing your working for each case, and explain why the final answer requires addition rather than multiplication. [6 marks]

Count separately: (2 flags: 5×4), (3 flags: 5×4×3), (4 flags: 5×4×3×2), (5 flags: 5×4×3×2×1). Use multiplication within each case (events in sequence), then add all cases because they are mutually exclusive events (a signal uses either 2 OR 3 OR 4 OR 5 flags, not simultaneously). Final sum = 20 + 60 + 120 + 120 = 320.

Assertion-Reason Questions for Permutations and Combinations

Q1. Assertion (A):

If an event can occur in m ways and a second event in n ways, the total number of occurrences of both events in order is m + n.

Reason (R):

The Fundamental Principle of Counting states that total occurrences equal the product of individual event outcomes.

Answer: D — A is false but R is true — Both the assertion and reason are false. The assertion incorrectly uses addition instead of multiplication. The Fundamental Principle uses multiplication (m × n), not addition. The reason correctly states the principle, but it contradicts the false assertion.

Q2. Assertion (A):

The number of 3-digit numbers formed from digits 1, 2, 3, 4, 5 without repetition is 60.

Reason (R):

We have 5 choices for the first digit, 4 for the second, and 3 for the third, so 5 × 4 × 3 = 60.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why the assertion is true. The calculation 5 × 4 × 3 = 60 directly justifies the number of 3-digit numbers without repetition.

Q3. Assertion (A):

The number of 4-letter codes from 26 letters with repetition allowed is 26^4.

Reason (R):

Each of the 4 positions can be filled independently in 26 ways because repetition is allowed.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason provides the correct justification. Since repetition is allowed, each position has 26 independent choices, giving 26 × 26 × 26 × 26 = 26^4 codes.

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Case-Based Questions (CBQ) for Permutations and Combinations

Case Study:

A school is organizing a science fair where students need to create display stands using different colored boards and decorative items. Each display must have: 1 colored board (available in 5 colors), 1 decorative item (available in 3 types), and 1 special effect (available in 2 types). The school wants to ensure that no two adjacent display stands have the same combination of colors and items. However, for the initial arrangement of the first 10 display stands placed in a row, this adjacency rule is not yet enforced. Calculate how many different display combinations can be created for the first 10 stands.

Q1. How many different display combinations are possible for a single stand?

  • A. 10
  • B. 30 ✓
  • C. 15
  • D. 5

Answer: B

Q2. If all 10 display stands must have different combinations, how many ways can they be arranged?

  • A. 30 × 29 × 28 × ... × 21 ✓
  • B. 30!/(20!)
  • C. 10!
  • D. 30^10

Answer: A

Q3. How many ways can 3 stands be selected from 10 stands to display red boards?

  • A. 10 × 9 × 8 ✓
  • B. 120
  • C. 3!
  • D. 30

Answer: A

Q4. If repetition of combinations is allowed across all 10 stands, how many total arrangements exist?

  • A. 10!
  • B. 30^10 ✓
  • C. 30 × 10
  • D. 300

Answer: B

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