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CBSE • Class XI • Mathematics • Ch 6
Estimated Time: 45 Mins
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Permutations and Combinations

In Class 11 Mathematics, "Permutations and Combinations" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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🔢 Have You Ever Wondered?

How many millions of distinct alphanumeric passwords can be generated from 8 characters, or in how many ways can a cricket selector pick a world-cup e...

How many millions of distinct alphanumeric passwords can be generated from 8 characters, or in how many ways can a cricket selector pick a world-cup eleven from 15 players? Combinatorics is the mathematical science of counting without listing.

Why This Chapter Matters

In Class 11 Mathematics, "Permutations and Combinations" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Factorial notation $n!$.
  • Fundamental principle of counting.
  • Algebraic equations.

What You Will Learn (Core Objectives)

  • Apply Fundamental Principle of Multiplication and Addition.
  • Define Factorial $n!$ and compute permutations: $^nP_r = \frac{n!}{(n-r)!}$.
  • Compute permutations with repetition: $\frac{n!}{p_1! p_2! \dots p_k!}$.
  • Define and apply Combinations: $^nC_r = \frac{n!}{r!(n-r)!}$.
  • Apply combinatorial identities: $^nC_r = ^nC_{n-r}$ and Pascal's rule $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.

Chapter Roadmap & Progression

1 1. Fundamental Principles & Factori...
2 2. Permutations: Order Matters
3 3. Combinations: Selection Only

Complete Concept Guide (100% Curriculum Coverage)

1. Fundamental Principles & Factorials

Multiplication Principle: If an event can occur in $m$ different ways, followed by another event in $n$ ways, both in succession occur in $\mathbf{m \times n}$ ways.
Factorial: $\mathbf{n! = n \times (n-1) \times \dots \times 1}$ with $0! = 1$.

2. Permutations: Order Matters

A Permutation is an arrangement in a definite order of a number of objects taken some or all at a time: $$\mathbf{^nP_r = \frac{n!}{(n-r)!}} \quad (0 \le r \le n)$$ When objects are not all distinct (e.g. letters in 'MISSISSIPPI'): $\frac{n!}{p!q!r!}$.

3. Combinations: Selection Only

A Combination is a selection of items where order is completely irrelevant: $$\mathbf{^nC_r = \frac{n!}{r!(n-r)!}} \quad \text{and} \quad \mathbf{^nP_r = r! \cdot ^nC_r}$$ Key identities: $^nC_a = ^nC_b \implies a = b \text{ or } a + b = n$; and $\mathbf{^nC_r + ^nC_{r-1} = ^{n+1}C_r}$.

Visual Learning & Conceptual Map

Permutations and Combinations Master Matrix

Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture

1. Fundamental Principles & Factorials • 2. Permutations: Order Matters

Chapter Summary & 10 Key Takeaways

Takeaway 1
Order Sensitivity: Permutations prioritize sequence; Combinations prioritize group membership.
Takeaway 2
Factorial Base: $0! = 1$ by combinatorial definition.
Takeaway 3
Permutation with Repeats: Dividing total factorials by individual frequency factorials.
Takeaway 4
Complementary Combination: $^nC_r = ^nC_{n-r}$ halving computation time.
Takeaway 5
Pascal's Identity: Recursive formula underlying Pascal's Triangle.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
In how many ways can the letters of the word 'PERMUTATIONS' be arranged if words start with P and end with S?
Reveal Answer & Explanation
Answer: Total letters = 12 (T repeats twice). Fixing P at start and S at end leaves 10 letters with 2 T's. Total ways $= \frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400$.
1,814,400 ways.
2
If $^{n}C_9 = ^{n}C_8$, find $^{n}C_{17}$.
Reveal Answer & Explanation
Answer: $^{n}C_a = ^{n}C_b \implies n = a + b = 9 + 8 = 17$. Therefore, $^{17}C_{17} = 1$.
n = 17, so 17C17 = 1.
3
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 exactly 3 girls?
Reveal Answer & Explanation
Answer: Choose 3 girls from 4 in $^4C_3$ ways, and remaining 4 boys from 9 in $^9C_4$ ways. Total ways $= ^4C_3 \times ^9C_4 = 4 \times 126 = 504$.
504 ways.
4
Evaluate $\frac{8!}{6! \times 2!}$.
Reveal Answer & Explanation
Answer: $\frac{8 \times 7 \times 6!}{6! \times 2} = \frac{56}{2} = 28$.
28.
5
How many 3-digit even numbers can be formed using digits 1, 2, 3, 4, 6, 7 if no digit is repeated?
Reveal Answer & Explanation
Answer: Units place must be even: 2, 4, or 6 (3 choices). Tens place: remaining 5 choices. Hundreds place: remaining 4 choices. Total numbers $= 4 \times 5 \times 3 = 60$.
60 numbers.
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