Chapter 6 — PERMUTATIONS AND COMBINATIONS

Theory

This is a selection + arrangement problem.

Strategy:
• First select required letters
• Then arrange them

\[ \text{Total} = \text{Selection} \times \text{Arrangement} \]

Word: DAUGHTER
Vowels: A, U, E (3)
Consonants: D, G, H, T, R (5)

Solution Roadmap

Step 1: Choose 2 vowels from 3
Step 2: Choose 3 consonants from 5
Step 3: Arrange selected 5 letters
Step 4: Multiply

Solution

Choose 2 vowels from 3:

\[ ^3C_2 = 3 \]

Choose 3 consonants from 5:

\[ ^5C_3 = 10 \]

Total selections:

\[ 3 \times 10 = 30 \]

Arrange 5 selected letters:

\[ 5! = 120 \]

Total words:

\[ \begin{aligned} \text{Total} &= 30 \times 5! \ &= 30 \times 120 \ &= 3600 \end{aligned} \]

Final Answer

\(3600\)

Exam Significance

This is a very important hybrid problem.

• Combines combination + permutation
• Frequently asked in CBSE exams
• Core pattern for JEE word problems

Key takeaway: Select first, then arrange

Theory

This is a grouping (block) arrangement problem.

When elements must stay together:
• Treat them as a single block
• Arrange blocks first
• Then arrange inside each block

\[ \text{Total} = \text{Block arrangement} \times \text{Internal arrangements} \]

Word: EQUATION
Vowels: A, E, I, O, U (5)
Consonants: Q, T, N (3)

Solution Roadmap

Step 1: Group vowels as one block
Step 2: Group consonants as one block
Step 3: Arrange the two blocks
Step 4: Arrange within each block
Step 5: Multiply

Solution

Treat vowels as one block and consonants as another block.

Arrange the two blocks:

\[ 2! \]

Arrange vowels within their block:

\[ 5! \]

Arrange consonants within their block:

\[ 3! \]

Total number of words:

\[ \begin{aligned} \text{Total} &= 2! \times 5! \times 3! \\ &= 2 \times 120 \times 6 \\ &= 1440 \end{aligned} \]

Final Answer

\(1440\)

Exam Significance

This is a classic grouping (block method) problem.

• Very common in CBSE board exams
• Extremely important for JEE permutations
• Foundation for “together / apart” problems

Key takeaway: When items must stay together → treat as one block

Theory

This is a case-based selection problem.

Key interpretations:
• Exactly k → single case
• At least k → k, k+1, ...
• At most k → 0 to k

\[ \text{Total} = \sum (\text{valid cases}) \]

Total: 9 boys, 4 girls → committee size = 7

Solution Roadmap

Step 1: Break into cases
Step 2: Select girls
Step 3: Select remaining boys
Step 4: Add all valid cases

Solution

(i) Exactly 3 girls

\[ \begin{aligned} ^4C_3 \times ^9C_4 &= 4 \times 126 \\ &= 504 \end{aligned} \]

(ii) At least 3 girls

Cases: 3 girls or 4 girls

\[ \begin{aligned} &= ^4C_3 \cdot ^9C_4 + ^4C_4 \cdot ^9C_3 \\ &= 4 \times 126 + 1 \times 84 \\ &= 504 + 84 \\ &= 588 \end{aligned} \]

(iii) At most 3 girls

Cases: 0, 1, 2, 3 girls

\[ \begin{aligned} &= ^4C_0 \cdot ^9C_7 + ^4C_1 \cdot ^9C_6 + ^4C_2 \cdot ^9C_5 + ^4C_3 \cdot ^9C_4 \\ &= 36 + 336 + 756 + 504 \\ &= 1632 \end{aligned} \]

Final Answer

(i) \(504\)
(ii) \(588\)
(iii) \(1632\)

Exam Significance

This is a high-frequency case-based selection problem.

• Very common in CBSE board exams
• Critical for JEE combinatorics
• Tests interpretation of “at least / at most”

Key takeaway: Convert words into cases, then sum valid combinations

Theory

This is a dictionary (lexicographic order) problem.

Key idea:
Words are arranged alphabetically.
All words starting with letters before E will come first.

Letters in EXAMINATION:
A(2), I(2), N(2), and others single

\[ \text{Permutations with repetition} = \frac{n!}{p!q!r!} \]

Solution Roadmap

Step 1: Identify letters before E
Step 2: Fix such letter at first position
Step 3: Arrange remaining letters
Step 4: Use repetition formula

Solution

Only letter before E in given word = A

So, count all words starting with A.

Fix A in first position.

Remaining letters: E, X, A, M, I, I, N, N, T, O → total 10 letters
Repetitions: I(2), N(2)

\[ \begin{aligned} \text{Number of words} &= \frac{10!}{2! \cdot 2!} \\ &= \frac{3628800}{4} \\ &= 907200 \end{aligned} \]

Final Answer

\(907200\)

Exam Significance

This is a high-level lexicographic permutation problem.

• Important for JEE Advanced
• Tests ordering + permutation with repetition
• Used in ranking and dictionary problems

Key takeaway: For dictionary problems → fix first letter and count permutations

Theory

This is a digit arrangement with restriction.

Divisibility rule:
A number divisible by 10 must end in 0.

Strategy:
• Fix the constrained position first
• Arrange remaining digits

\[ \text{Total ways} = \text{Arrangement of remaining digits} \]

Solution Roadmap

Step 1: Fix units digit (0)
Step 2: Arrange remaining 5 digits
Step 3: Use permutation (all distinct)

Solution

Units digit must be 0 → fixed.

Remaining digits: 1, 3, 5, 7, 9 (5 digits)

These 5 digits can be arranged in:

\[ 5! = 120 \]

Hence, total numbers:

\[ 120 \]

Final Answer

\(120\)

Exam Significance

This is a restriction-based permutation problem.

• Very common in CBSE exams
• Important for JEE digit problems
• Tests divisibility + arrangement logic

Key takeaway: For digit constraints → fix position first, then arrange

Theory

This is a selection + arrangement problem.

Strategy:
• Select required letters (combinations)
• Arrange selected letters (permutations)

\[ \text{Total} = \text{Selection} \times \text{Arrangement} \]

Vowels = 5
Consonants = 21

Solution Roadmap

Step 1: Choose 2 vowels from 5
Step 2: Choose 2 consonants from 21
Step 3: Arrange 4 selected letters
Step 4: Multiply

Vowels → \(^5C_2\)
Consonants → \(^{21}C_2\)
Arrange → \(4!\)
Multiply all

Solution

Choose 2 vowels:

\[ ^5C_2 = 10 \]

Choose 2 consonants:

\[ ^{21}C_2 = 210 \]

Arrange the 4 selected letters:

\[ 4! = 24 \]

Total number of words:

\[ \begin{aligned} \text{Total} &= ^5C_2 \times ^{21}C_2 \times 4! \\ &= 10 \times 210 \times 24 \\ &= 50400 \end{aligned} \]

Final Answer

\(50400\)

Exam Significance

This is a fundamental hybrid combinatorics problem.

• Very common in CBSE board exams
• Important for JEE mixed problems
• Tests selection + permutation together

Key takeaway: Choose first, then arrange

Theory

This is a distribution-based selection problem.

When constraints like “at least k” are given:
• Break into valid cases
• Solve each case separately
• Add results

\[ \text{Total} = \sum (\text{valid distributions}) \]

Solution Roadmap

Step 1: Let questions from Part I = x, Part II = y
Step 2: Use x + y = 8 with x ≥ 3, y ≥ 3
Step 3: Find valid pairs
Step 4: Compute each case and add

Solution

Valid distributions: \((3,5), (4,4), (5,3)\)

Case 1: (3,5)

\[ ^5C_3 \cdot ^7C_5 = 10 \times 21 = 210 \]

Case 2: (4,4)

\[ ^5C_4 \cdot ^7C_4 = 5 \times 35 = 175 \]

Case 3: (5,3)

\[ ^5C_5 \cdot ^7C_3 = 1 \times 35 = 35 \]

Total ways:

\[ \begin{aligned} \text{Total} &= 210 + 175 + 35 \\ &= 420 \end{aligned} \]

Final Answer

\(420\)

Exam Significance

This is a classic distribution + selection problem.

• Very common in CBSE exams
• Important for JEE combinatorics
• Tests interpretation of constraints

Key takeaway: Convert constraints into valid cases, then sum

Theory

This is a conditional selection problem.

Deck composition:
• Kings = 4
• Non-kings = 48

For “exactly one king”:
• Select required king(s)
• Select remaining from non-kings

\[ \text{Total} = ^4C_1 \times ^{48}C_4 \]

Solution Roadmap

Step 1: Choose 1 king from 4
Step 2: Choose 4 non-kings from 48
Step 3: Multiply selections

Solution

Choose 1 king:

\[ ^4C_1 = 4 \]

Choose 4 non-kings:

\[ ^{48}C_4 = \frac{48 \times 47 \times 46 \times 45}{24} = 194580 \]

Total combinations:

\[ \begin{aligned} \text{Total} &= ^4C_1 \times ^{48}C_4 \\ &= 4 \times 194580 \\ &= 778320 \end{aligned} \]

Final Answer

\(778320\)

Exam Significance

This is a standard card-combination problem.

• Very important for probability chapter
• Common in JEE and board exams
• Tests “exactly k items” logic

Key takeaway: “Exactly k” ⇒ choose required first, then remaining

Theory

This is a position-restriction permutation problem.

Strategy:
• Fix positions based on restriction
• Arrange individuals within those positions

Total seats = 9 → positions: 1 to 9
Even positions: 2, 4, 6, 8 (4 places)

Solution Roadmap

Step 1: Fix women in even positions
Step 2: Place men in remaining positions
Step 3: Arrange each group independently
Step 4: Multiply

Solution

Even positions (2, 4, 6, 8) are fixed for women.

Number of ways to arrange 4 women:

\[ 4! \]

Remaining 5 positions (odd positions) are for men.

Number of ways to arrange 5 men:

\[ 5! \]

Total arrangements:

\[ \begin{aligned} \text{Total} &= 4! \times 5! \\ &= 24 \times 120 \\ &= 2880 \end{aligned} \]

Final Answer

\(2880\)

Exam Significance

This is a classic position-restriction problem.

• Very common in CBSE exams
• Important for JEE seating arrangement problems
• Tests fixing positions + permutation

Key takeaway: Fix restricted positions first, then arrange remaining

Theory

This is a conditional grouping problem.

When a group must act together:
• Treat them as a unit
• Consider separate cases

Condition:
• Either all 3 selected OR none selected

\[ \text{Total} = \text{Case 1} + \text{Case 2} \]

Solution Roadmap

Step 1: Case 1 → all 3 included
Step 2: Case 2 → none included
Step 3: Compute each case
Step 4: Add results

Solution

Case 1: All 3 students included

Remaining students to choose = \(10 - 3 = 7\)
Available students = \(25 - 3 = 22\)

\[ ^{22}C_7 \]

Case 2: None of the 3 students included

Choose all 10 students from remaining 22:

\[ ^{22}C_{10} \]

Total number of ways:

\[ \begin{aligned} \text{Total} &= ^{22}C_7 + ^{22}C_{10} \end{aligned} \]

Final Answer

\(^{22}C_7 + ^{22}C_{10}\)

Exam Significance

This is a group constraint + case-based selection problem.

• Very important for JEE combinatorics
• Tests grouping and logical conditions
• Common in advanced selection problems

Key takeaway: When elements act together → split into cases (all or none)

Theory

This is a grouping + permutation with repetition problem.

When identical elements must stay together:
• Treat them as one block
• Then apply permutation with repetition

Letter count:
A = 3, S = 4, I = 2, N = 2, T = 1, O = 1

\[ \text{Permutations} = \frac{n!}{p!q!r!} \]

Solution Roadmap

Step 1: Group all S’s as one block
Step 2: Count total units
Step 3: Apply permutation with repetition
Step 4: Simplify

Solution

Treat all 4 S’s as one block (SSSS).

Remaining letters: 3 A’s, 2 I’s, 2 N’s, T, O

Total units:

\[ 1 + 3 + 2 + 2 + 1 + 1 = 10 \]

Apply permutation with repetition:

\[ \frac{10!}{3! \cdot 2! \cdot 2!} \]

Simplifying:

\[ \begin{aligned} \frac{10!}{3!2!2!} &= \frac{3628800}{6 \cdot 2 \cdot 2} \\ &= \frac{3628800}{24} \\ &= 151200 \end{aligned} \]

Final Answer

\(151200\)

Exam Significance

This is a high-value permutation with repetition problem.

• Very common in CBSE boards
• Extremely important for JEE Advanced
• Tests grouping + repetition simultaneously

Key takeaway: When identical items must stay together → treat as one block, then apply repetition formula