Chapter 6 — PERMUTATIONS AND COMBINATIONS

Theory

When order matters and repetition is not allowed, we use permutations.

\[ ^nP_r = \frac{n!}{(n-r)!} \]

A 3-digit number has positions: Hundreds, Tens, Units, and each position must be filled with a distinct digit.

Solution Roadmap

Step 1: Total digits available → 9 (1 to 9)
Step 2: Number of positions → 3
Step 3: No repetition ⇒ decreasing choices
Step 4: Apply FPC or permutation formula

Solution

Total digits = 9

First position: 9 choices
Second position: 8 choices
Third position: 7 choices

\[ \begin{aligned} \text{Total numbers} &= 9 \times 8 \times 7 \\ &= 504 \end{aligned} \]

\[ ^9P_3 = 504 \]

Final Answer

\(504\)

Exam Significance

This is a standard permutation model problem.

• Very common in CBSE board exams
• Forms base for all digit arrangement problems
• Essential for JEE questions with restrictions

Key takeaway: No repetition + order matters ⇒ use permutation (decreasing choices)

Theory

A 4-digit number ranges from 1000 to 9999, so the first digit cannot be zero.

Since no repetition is allowed and order matters, this is a permutation with restriction.

Strategy: Fix the first digit carefully, then apply permutation to remaining positions.

Solution Roadmap

Step 1: Identify restriction → first digit ≠ 0
Step 2: Count choices for first digit
Step 3: Fill remaining positions without repetition
Step 4: Multiply using FPC

Solution

Total digits available: \(0,1,2,\ldots,9\) ⇒ 10 digits

First digit (thousands place):
Cannot be 0 ⇒ 9 choices (1–9)

Remaining three digits:
After fixing first digit, 9 digits remain (including 0)
Fill 3 places without repetition:

\[ {}^{9}P_3 = 9 \times 8 \times 7 \]

Total numbers:

\[ \begin{aligned} \text{Total} &= 9 \times {}^{9}P_3 \\ &= 9 \times 9 \times 8 \times 7 \\ &= 4536 \end{aligned} \]

Final Answer

\(4536\)

Exam Significance

This is a classic restriction-based permutation problem.

• Very common in CBSE board exams
• Tests understanding of leading digit constraint
• Highly important for JEE digit-formation problems

Key takeaway: Always handle leading digit restrictions before applying permutations

Theory

A number is even if its unit digit is even.

Since order matters and no repetition is allowed, we use permutation under constraint.

Strategy: Fix the constrained position first (units place), then fill remaining positions.

Solution Roadmap

Step 1: Identify condition → last digit must be even
Step 2: Choose unit digit from even digits
Step 3: Fill remaining positions without repetition
Step 4: Multiply using FPC

Solution

Given digits: \(1,2,3,4,6,7\)

Step 1: Unit place (even condition)
Even digits = \(2,4,6\) ⇒ 3 choices

Step 2: Remaining digits
After fixing unit digit, 5 digits remain

Step 3: Fill hundreds and tens
Hundreds place: 5 choices
Tens place: 4 choices

\[ \begin{aligned} \text{Total numbers} &= 5 \times 4 \times 3 \\ &= 60 \end{aligned} \]

\[ = {}^{5}P_{2} \times 3 \]

Final Answer

\(60\)

Exam Significance

This is a constraint-first permutation problem, very important in exams.

• Common in CBSE board exams
• Core model for JEE digit-based problems
• Teaches position-based restriction handling

Key takeaway: Always fix the restricted position first (here, units place)

Theory

This problem has two parts: Total arrangements and arrangements under a condition (even numbers).

Since order matters and no repetition is allowed, we use permutations.

For even numbers, the unit digit must be even.

Solution Roadmap

Part 1: Count total 4-digit numbers using permutation
Part 2: Apply condition → last digit even
Part 3: Fix unit place first
Part 4: Arrange remaining digits

Solution

Part 1: Total 4-digit numbers

We arrange 4 digits out of 5 without repetition:

\[ \begin{aligned} {}^{5}P_{4} &= 5 \times 4 \times 3 \times 2 \\ &= 120 \end{aligned} \]

Total numbers = \(120\)

Part 2: Even numbers

Even digits available: \(2,4\) ⇒ 2 choices for unit place

After fixing unit digit, 4 digits remain

Fill remaining 3 positions:

\[ {}^{4}P_{3} = 4 \times 3 \times 2 \]

Total even numbers:

\[ \begin{aligned} 2 \times {}^{4}P_{3} &= 2 \times 4 \times 3 \times 2 \\ &= 48 \end{aligned} \]

Final Answer

Total numbers = \(120\)
Even numbers = \(48\)

Exam Significance

This is a two-layer counting problem: total vs restricted counting.

• Very common CBSE board question type
• Builds case-based thinking
• Important for JEE multi-condition problems

Key takeaway: First count total, then apply restriction by fixing the constrained position

Theory

When positions are different (chairman ≠ vice chairman), order matters.

Hence, this is a permutation problem, not combination.

Selecting two people for two distinct posts is equivalent to:

\[ ^8P_2 = \frac{8!}{6!} \]

Solution Roadmap

Step 1: Identify roles → Chairman and Vice Chairman
Step 2: Order matters → use permutation
Step 3: Select first position
Step 4: Select second position from remaining

Solution

Total persons = 8

Step 1: Choose Chairman
8 choices

Step 2: Choose Vice Chairman
Remaining persons = 7 ⇒ 7 choices

\[ \begin{aligned} \text{Total ways} &= 8 \times 7 \\ &= 56 \end{aligned} \]

\[ = {}^{8}P_{2} \]

Final Answer

\(56\)

Exam Significance

This is a classic role-based permutation problem.

• Very common confusion: combination vs permutation
• Direct CBSE board question type
• Highly important for JEE selection vs arrangement logic

Key takeaway: Different roles ⇒ order matters ⇒ use permutation

Theory

Permutation formula:

\[ ^nP_r = \frac{n!}{(n-r)!} \]

In ratio problems, avoid full expansion. Instead: convert into factorial form and cancel systematically.

Key identity:

\[ n! = n \cdot (n-1)! \]

Solution Roadmap

Step 1: Convert ratio into fraction
Step 2: Apply permutation formula
Step 3: Cancel common factorial terms
Step 4: Solve resulting simple equation

Solution

Given:

\[ \frac{{}^{\,n-1}P_3}{{}^{\,n}P_4} = \frac{1}{9} \]

Using permutation formula:

\[ \begin{aligned} \frac{\dfrac{(n-1)!}{(n-4)!}}{\dfrac{n!}{(n-4)!}} &= \frac{1}{9} \end{aligned} \]

Cancel \((n-4)!\):

\[ \frac{(n-1)!}{n!} = \frac{1}{9} \]

Using \(n! = n \cdot (n-1)!\):

\[ \begin{aligned} \frac{(n-1)!}{n \cdot (n-1)!} &= \frac{1}{9} \ \frac{1}{n} &= \frac{1}{9} \\ n &= 9 \end{aligned} \]

Final Answer

\(n = 9\)

Exam Significance

This is a high-value JEE pattern problem involving ratio simplification.

• Tests algebraic manipulation with factorials
• Common in competitive exams (especially JEE)
• Requires smart cancellation, not brute calculation

Key takeaway: In permutation ratios, always cancel factorials before solving

Theory

Permutation formula:

\[ ^nP_r = \frac{n!}{(n-r)!} \]

In equations involving permutations: convert into factorial form and reduce.

Also remember domain: \[ 0 \le r \le n \]

Solution Roadmap

Step 1: Convert permutation into factorial form
Step 2: Cancel common factorial terms
Step 3: Reduce to algebraic equation
Step 4: Solve and check validity of \(r\)

Solution

(i) \({}^{5}P_r = 2 \cdot {}^{6}P_{r-1}\)

\[ \begin{aligned} \frac{5!}{(5-r)!} &= 2 \cdot \frac{6!}{(7-r)!} \end{aligned} \]

Using \(6! = 6 \cdot 5!\) and simplifying:

\[ \begin{aligned} \frac{1}{(5-r)!} &= \frac{12}{(7-r)(6-r)(5-r)!} \\ (7-r)(6-r) &= 12 \\ r^2 - 13r + 30 &= 0 \\ (r-10)(r-3) &= 0 \\ r &= 10,\;3 \end{aligned} \]

Valid domain: \(0 \le r \le 5\) ⇒ \(r = 3\)

(ii) \({}^{5}P_r = {}^{6}P_{r-1}\)

\[ \begin{aligned} \frac{5!}{(5-r)!} &= \frac{6!}{(7-r)!} \end{aligned} \]

Simplifying:

\[ \begin{aligned} 1 &= \frac{6}{(7-r)(6-r)} \\ (7-r)(6-r) &= 6 \\ r^2 - 13r + 36 &= 0 \\ (r-9)(r-4) &= 0 \\ r &= 9,\;4 \end{aligned} \]

Valid domain ⇒ \(r = 4\)

Final Answer

(i) \(r = 3\)
(ii) \(r = 4\)

Exam Significance

This is a JEE-level algebraic permutation problem.

• Combines algebra + permutation
• Tests simplification speed
• Requires domain validation (very important)

Key takeaway: Solve algebraically, then filter using domain constraints

Theory

When all objects are distinct and each is used exactly once, the number of arrangements is:

\[ n! \]

This is a permutation of all distinct objects.

The word EQUATION contains 8 different letters, so total arrangements = \(8!\).

Solution Roadmap

Step 1: Count total letters
Step 2: Check if repetition exists → No
Step 3: Apply factorial (all distinct)
Step 4: Compute value

Solution

Total letters in EQUATION = 8 (all distinct)

\[ \begin{aligned} \text{Total words} &= 8! \\ &= 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \\ &= 40320 \end{aligned} \]

Final Answer

\(40320\)

Exam Significance

This is a basic permutation model for distinct objects.

• Very common CBSE board question
• Foundation for word arrangement problems
• Base case before introducing repeated letters

Key takeaway: All distinct objects ⇒ total arrangements = \(n!\)

Theory

The word MONDAY has 6 distinct letters.

Since order matters and no repetition, we use:

\[ ^nP_r = \frac{n!}{(n-r)!} \]

For conditions (like vowel at first position), fix the constrained position first.

Solution Roadmap

(i) Select and arrange 4 letters
(ii) Arrange all letters
(iii) Fix first position (vowel), then arrange remaining

Solution

(i) 4 letters used

\[ \begin{aligned} {}^{6}P_{4} &= \frac{6!}{2!} \\ &= 6 \times 5 \times 4 \times 3 \\ &= 360 \end{aligned} \]

(ii) All 6 letters used

\[ \begin{aligned} {}^{6}P_{6} &= 6! \\ &= 720 \end{aligned} \]

(iii) First letter is a vowel

Vowels: \(O, A\) ⇒ 2 choices

Remaining 5 letters arranged in 5! ways

\[ \begin{aligned} \text{Total} &= 2 \times 5! \\ &= 2 \times 120 \\ &= 240 \end{aligned} \]

Final Answer

(i) \(360\)
(ii) \(720\)
(iii) \(240\)

Exam Significance

This is a multi-case permutation problem.

• Tests concept switching (selection → arrangement → restriction)
• Common in CBSE exams (multi-part questions)
• Very important for JEE case-based reasoning

Key takeaway: For constraints, always fix that position first before arranging others

Theory

When letters repeat, total distinct permutations are:

\[ \frac{n!}{p!q!r!\dots} \]

For restriction like “not together”, use: Complement method

Required = Total arrangements − Arrangements where all I’s are together

Solution Roadmap

Step 1: Count total permutations with repetition
Step 2: Treat all I’s as one block
Step 3: Count arrangements with I’s together
Step 4: Subtract from total

Solution

Letters in MISSISSIPPI:
M = 1, I = 4, S = 4, P = 2 ⇒ total = 11 letters

Step 1: Total permutations

\[ \begin{aligned} \text{Total} &= \frac{11!}{4! \cdot 4! \cdot 2!} \\ &= 34650 \end{aligned} \]

Step 2: I’s together (treat as one block)

Objects: (IIII), M, S, S, S, S, P, P ⇒ 8 objects
Repetitions: S = 4, P = 2

\[ \begin{aligned} \text{Together cases} &= \frac{8!}{4! \cdot 2!} \\ &= 840 \end{aligned} \]

Step 3: Required (not together)

\[ \begin{aligned} \text{Required} &= 34650 - 840 \\ &= 33810 \end{aligned} \]

Final Answer

\(33810\)

Exam Significance

This is a high-level repeated permutation + restriction problem.

• Very important for JEE (frequent pattern)
• Tests complement method understanding
• Combines grouping + repetition handling

Key takeaway: For “not together” → use complement (Total − Together)

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Theory

Total letters = 12
Repetition: T appears 2 times
Vowels: A, E, I, O, U (5 vowels)

For repeated letters:

\[ \frac{n!}{p!q!} \]

For constraints: Fix positions → group elements → arrange remaining

Solution Roadmap

(i) Fix first & last positions
(ii) Treat vowels as one block
(iii) Fix relative positions (gap method)

Solution

(i) Starts with P and ends with S

Fix P at first and S at last ⇒ remaining 10 positions

Remaining letters include T repeated twice

\[ \begin{aligned} \text{Ways} &= \frac{10!}{2!} \\ &= 1814400 \end{aligned} \]

(ii) Vowels together

Treat vowels (AEIOU) as one block ⇒ total objects = 8
Repetition: T appears twice

\[ \begin{aligned} \text{Arrangements of blocks} &= \frac{8!}{2!} \end{aligned} \]

Arrange vowels internally:

\[ 5! \] \[ \begin{aligned} \text{Total} &= \frac{8!}{2!} \times 5! \\ &= 2419200 \end{aligned} \]

(iii) Exactly 4 letters between P and S

Distance between P and S = 5 positions

Possible position pairs: 7
Each pair has 2 arrangements (P-S or S-P)

\[ 7 \times 2 = 14 \]

Remaining 10 letters arranged:

\[ \frac{10!}{2!} \] \[ \begin{aligned} \text{Total} &= 14 \times \frac{10!}{2!} \\ &= 25401600 \end{aligned} \]

Final Answer

(i) \(1814400\)
(ii) \(2419200\)
(iii) \(25401600\)

Exam Significance

This is a complete mixed-constraint permutation problem.

• Tests fixing, grouping, and positioning simultaneously
• Very important for JEE Advanced problems
• Combines multiple core concepts in one question

Key takeaway: Master 3 tools → Fix positions, Group elements, Control gaps