NCERT Chapter 6

Permutations & Combinations

Learn the art of counting without listing: from the fundamental counting principle and factorials to arrangements and selections.

~50 min read 6–8 marks in exam kemh106.pdf

What you will learn in this chapter

  1. Fundamental Principles of Counting (Multiplication & Addition)
  2. Factorial Notation & Why $0! = 1$
  3. Permutations ($^nP_r$) — When Order Matters
  4. Permutations of Identical Objects (Multiset Formula)
  5. Key Arrangement Strategies: Tie-Together & Gap Methods
  6. Combinations ($^nC_r$) — When Order Does NOT Matter
  7. Core Identities of $^nC_r$ & Geometric Applications
  8. Worked Examples — Easy to Hard
  9. NCERT Key Exercise Questions Solved
  10. Common Exam Traps & Quick Self-Check Quiz

1. Fundamental Principles of Counting

The Multiplication Principle (AND)

If an event can occur in $m$ different ways, following which a second event can occur in $n$ different ways, then the total number of ways both events can occur in sequence is:

$$\mathbf{m \times n}$$

Example: You have 3 shirts and 2 pants. Total outfits $= 3 \times 2 = 6$.

The Addition Principle (OR)

If event $A$ can occur in $m$ ways and event $B$ can occur in $n$ ways, and they cannot occur simultaneously (mutually exclusive), then either $A$ OR $B$ can occur in:

$$\mathbf{m + n} \quad \text{ways}$$

Example: You can choose 1 coffee out of 4 kinds OR 1 tea out of 3 kinds $= 4 + 3 = 7$ choices.

Repetition Allowed vs Repetition Not Allowed

Consider forming a 3-digit number using digits $\{1, 2, 3, 4, 5\}$:

2. Factorial Notation & Why $0! = 1$

Definition: Factorial Notation ($n!$)

For any positive integer $n$, the product of the first $n$ natural numbers is denoted by $n!$ (read "$n$ factorial"):

$$n! = n \times (n - 1) \times (n - 2) \times \dots \times 3 \times 2 \times 1$$

Recursive property: $n! = n \times (n - 1)!$.

Values of First Few Factorials (Memorize These!)

$$0! = 1, \quad 1! = 1, \quad 2! = 2, \quad 3! = 6, \quad 4! = 24, \quad 5! = 120, \quad 6! = 720, \quad 7! = 5040$$

Why is $0! = 1$?

Using the identity $(n - 1)! = \frac{n!}{n}$, plug in $n = 1$:

$$(1 - 1)! = \frac{1!}{1} \implies 0! = \frac{1}{1} = 1$$

Intuitive reason: There is exactly 1 way to arrange zero objects: leave the space empty!

3. Permutations ($^nP_r$) — Order Matters!

Definition of Permutation

A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.

The number of permutations of $n$ distinct objects taken $r$ at a time ($0 \le r \le n$) without repetition is:

$$^nP_r = \frac{n!}{(n - r)!} = n(n - 1)(n - 2)\dots(n - r + 1)$$

Special values: $^nP_n = n!$, $^nP_0 = 1$, $^nP_1 = n$.

4. Permutations of Objects Not All Distinct

The Multiset Permutation Formula

The number of permutations of $n$ objects, where $p_1$ objects are of one kind, $p_2$ are of second kind, ..., and $p_k$ are of $k$-th kind is:

$$\mathbf{\frac{n!}{p_1! \cdot p_2! \dots p_k!}}$$

Classic Example (ROOT): The word ROOT has 4 letters where O repeats 2 times. Total arrangements $= \frac{4!}{2!} = \frac{24}{2} = 12$.

Classic Example (MISSISSIPPI): 11 letters total (1 M, 4 I, 4 S, 2 P). Total ways $= \frac{11!}{4! \cdot 4! \cdot 2!} = 34,650$.

5. Master Arrangement Strategies

Strategy A: The "Tie-Together" (String) Method

Use when certain objects must always be together.

  1. Tie the required items into a single bundle/block.
  2. Treat that block as 1 giant item. Count permutations of the items including the block.
  3. Multiply by the internal permutations within the block!
Strategy B: The "Gap" Method

Use when certain objects must NEVER be together.

  1. Arrange the unrestricted items first (say $k$ items in $k!$ ways).
  2. These $k$ items create $k + 1$ available "gaps" (spaces before, between, and after them).
  3. Place the restricted items into these gaps using combinations/permutations. No two will ever sit together!

6. Combinations ($^nC_r$) — Order Does NOT Matter

Permutation vs Combination

Think about a cricket team: Choosing 11 players out of 15 is a Combination (it doesn't matter who was picked first). But assigning batting positions 1 to 11 is a Permutation (order changes the lineup completely!).

Definition of Combination

The number of combinations (selections) of $n$ different objects taken $r$ at a time is:

$$^nC_r = \frac{n!}{r!(n - r)!} = \frac{^nP_r}{r!}$$

7. Core Properties of $^nC_r$ & Geometric Formulas

Properties of Combinations

Number of Diagonals in an $n$-gon

Selecting any 2 vertices gives a line segment ($^nC_2$). Subtracting the $n$ perimeter sides gives the diagonals:

$$\text{Diagonals} = ^nC_2 - n = \frac{n(n - 3)}{2}$$

Collinear Points & Lines / Triangles

From $n$ points where $m$ are collinear:

$$\text{Lines} = ^nC_2 - ^mC_2 + 1$$

$$\text{Triangles} = ^nC_3 - ^mC_3$$

8. Worked Examples — Easy to Hard

Example 1: Factorial Equation Easy
If $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$, find $x$.
Step 1: Express factorials in terms of the smallest factorial ($6!$): $$\frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{8 \times 7 \times 6!}$$
Step 2: Multiply both sides by $6!$: $$1 + \frac{1}{7} = \frac{x}{56} \implies \frac{8}{7} = \frac{x}{56}$$
Step 3: Solve for $x$: $$x = \frac{8 \times 56}{7} = 8 \times 8 = 64$$
Example 2: Words Formed with Repeating Letters Medium
How many different words (with or without meaning) can be formed using all letters of the word MATHEMATICS?
Step 1: Count letter frequencies: Total letters $n = 11$. Frequencies: M appears 2 times, A appears 2 times, T appears 2 times, H: 1, E: 1, I: 1, C: 1, S: 1.
Step 2: Apply multiset formula: $$\text{Total words} = \frac{11!}{2! \cdot 2! \cdot 2!} = \frac{39,916,800}{8} = 4,989,600$$
Example 3: Vowels Always Together (Tie-Together Method) Medium
Find the number of arrangements of the letters of the word DAUGHTER such that all vowels are always together.
Step 1: Identify vowels and consonants: Word has 8 letters: Vowels are $\{A, U, E\}$ (3 vowels). Consonants are $\{D, G, H, T, R\}$ (5 consonants).
Step 2: Tie vowels into a single block $[AUE]$: Now we have 5 consonants + 1 block $= 6$ objects. Number of ways to arrange these 6 objects $= 6! = 720$.
Step 3: Arrange vowels inside the block: The 3 vowels can be arranged among themselves in $3! = 6$ ways.
Total ways: $$6! \times 3! = 720 \times 6 = 4320$$
Example 4: Vowels Never Together (Gap Method) Hard
In how many ways can the letters of the word DAUGHTER be arranged so that no two vowels are ever together?
Step 1: Arrange the 5 consonants first: $$\_ D \_ G \_ H \_ T \_ R \_$$ Number of ways to arrange 5 consonants $= 5! = 120$.
Step 2: Count available gaps for vowels: 5 consonants create $5 + 1 = 6$ gaps (marked with $\_$).
Step 3: Place 3 vowels into the 6 gaps: Number of ways $= ^6P_3 = 6 \times 5 \times 4 = 120$.
Total ways: $$5! \times ^6P_3 = 120 \times 120 = 14,400$$
Example 5: Committee Selection with "At Least" Condition Hard
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of at least 3 girls?
Break into mutually exclusive cases: Since there are 4 girls available, "at least 3 girls" means either 3 girls or 4 girls:
  • Case 1 (3 girls and 4 boys): $^4C_3 \times ^9C_4 = 4 \times 126 = 504$.
  • Case 2 (4 girls and 3 boys): $^4C_4 \times ^9C_3 = 1 \times 84 = 84$.
Total combinations (Addition Principle): $$504 + 84 = 588$$

9. NCERT Key Exercise Questions Solved

NCERT Exercise 6.3 — Question 8

Q: How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
View Full Solution

There are 9 available digits $\{1, 2, ..., 9\}$ and we must choose and arrange 4 distinct digits.

$$^9P_4 = 9 \times 8 \times 7 \times 6 = 3024$$

NCERT Exercise 6.4 — Question 2

Q: If $^{2n}C_3 : ^nC_3 = 12 : 1$, find $n$.
View Full Solution

$$\frac{^{2n}C_3}{^nC_3} = \frac{\frac{2n(2n-1)(2n-2)}{3!}}{\frac{n(n-1)(n-2)}{3!}} = \frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = \frac{4(2n-1)}{n - 2} = 12$$

$$4(2n - 1) = 12(n - 2) \implies 8n - 4 = 12n - 24 \implies 4n = 20 \implies n = 5$$

NCERT Miscellaneous Exercise — Question 7

Q: In an examination, a question paper consists of 12 questions divided into two parts, Part I and Part II, containing 5 and 7 questions respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?
View Full Solution

Possible distributions of 8 questions between Part I (5 questions) and Part II (7 questions) with $\ge 3$ from each:

  • Case 1: 3 from Part I, 5 from Part II: $^5C_3 \times ^7C_5 = 10 \times 21 = 210$.
  • Case 2: 4 from Part I, 4 from Part II: $^5C_4 \times ^7C_4 = 5 \times 35 = 175$.
  • Case 3: 5 from Part I, 3 from Part II: $^5C_5 \times ^7C_3 = 1 \times 35 = 35$.

Total ways $= 210 + 175 + 35 = 420$.

10. Common Exam Traps & Quick Self-Check Quiz

Trap 1: Confusing Permutation with Combination

Ask yourself: Does changing the order create a brand new outcome? If yes $\to$ Permutation ($^nP_r$). If no $\to$ Combination ($^nC_r$). Picking a delegation is $^nC_r$; appointing President, Secretary is $^nP_r$.

Trap 2: Forgetting Internal Arrangements in Tie-Together

Students tie vowels $[A, E, I]$ together and count $6!$, but completely forget to multiply by $3!$ for their internal arrangements! Always multiply by the permutations within each tied block.

Q1: If $^nC_{10} = ^nC_{12}$, what is the value of $n$?

Answer
Using $^nC_a = ^nC_b \implies a + b = n$, we get $n = 10 + 12 = 22$.

Q2: How many diagonals does a regular decagon (10-sided polygon) have?

Answer
$\text{Diagonals} = ^{10}C_2 - 10 = \frac{10 \times 9}{2} - 10 = 45 - 10 = 35$.

Q3: What is $^nC_r + ^nC_{r-1}$ equal to?

Answer
$^{n+1}C_r$ (Pascal's rule).

Q4: In how many ways can 5 people sit on 5 chairs?

Answer
$5! = 120$ ways.

Q5: How many 3-letter words can be formed from letters of the word BIRD without repetition?

Answer
Word has 4 distinct letters. $^4P_3 = 4 \times 3 \times 2 = 24$.
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