Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
Combinatorics answers "how many ways" questions — arrangements (permutations) and selections (combinations). Mathematical Induction is a proof technique that establishes a result for all positive integers. Both topics appear regularly in board exams and entrance tests.
Fundamental Principle of Counting
- Multiplication Principle: If event A can occur in m ways and event B in n ways, both together can occur in m × n ways.
- Addition Principle: If event A can occur in m ways OR event B in n ways (mutually exclusive), either can occur in m + n ways.
Factorials
n! = n × (n−1) × (n−2) × … × 2 × 1
Special values: 0! = 1, 1! = 1
Permutations (Order Matters)
nPr = number of arrangements of r objects from n distinct objects
$$nPr = \frac{n!}{(n-r)!}$$
| Case | Formula |
|---|---|
| All n objects | n! |
| r from n (no repetition) | n!/(n−r)! |
| With repetition allowed | nʳ |
| With identical objects (p of one kind, q of another) | n!/(p! q!) |
| Circular arrangement | (n−1)! |
| Circular with two sides same (necklace) | (n−1)!/2 |
Combinations (Order Does Not Matter)
nCr = number of ways to choose r objects from n
$$nCr = \frac{n!}{r!(n-r)!}$$
Key Properties
| Property | Statement |
|---|---|
| Symmetry | nCr = nC(n−r) |
| Identity | nC0 = nCn = 1 |
| Pascal's rule | nCr = (n−1)C(r−1) + (n−1)Cr |
| Sum | nC0 + nC1 + … + nCn = 2ⁿ |
Mathematical Induction
Prove P(n) is true for all n ∈ ℕ:
1. Base step: Show P(1) is true. 2. Inductive hypothesis: Assume P(k) is true. 3. Inductive step: Prove P(k+1) is true using P(k). 4. Conclusion: By PMI, P(n) is true for all n ≥ 1.
Worked Examples
Example 1: In how many ways can 5 people sit in a row? In a circle? Row: 5! = 120. Circle: (5−1)! = 4! = 24.
Example 2: A committee of 3 is chosen from 8 people. How many ways? 8C3 = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 56.
Example 3: Prove by induction: 1 + 2 + 3 + … + n = n(n+1)/2.
∴ True for all n ∈ ℕ.
- P(1): LHS = 1; RHS = 1(2)/2 = 1. ✓
- Assume P(k): 1 + 2 + … + k = k(k+1)/2.
- P(k+1): LHS = k(k+1)/2 + (k+1) = (k+1)(k+2)/2 = RHS. ✓
Example 4: How many 3-digit numbers can be formed from digits 1–7 with no repetition? 7P3 = 7!/(7−3)! = 7 × 6 × 5 = 210.
Important Questions
One Mark
- Find 6P3 and 6C3.
- How many ways can 4 books be arranged on a shelf?
- State the base step of mathematical induction.
- Find the value of 10C7.
Short Answer (3 marks)
- In how many ways can a team of 4 boys and 3 girls be chosen from 8 boys and 5 girls?
- In how many ways can the letters of TRIANGLE be arranged so that vowels always come together?
- Prove by induction: 2 + 4 + 6 + … + 2n = n(n+1).
Long Answer (5 marks)
- Prove by mathematical induction: 1² + 2² + 3² + … + n² = n(n+1)(2n+1)/6.
- In an exam, a student must answer 7 out of 10 questions, with at least 3 from the first 5 and at least 2 from the last 5. Find the number of ways.
Common Mistakes
- Using nPr when combination (nCr) is required — check if order matters.
- Forgetting 0! = 1, which is needed in nC0 = 1 and nCn = 1.
- Mathematical induction: writing "assume P(n) is true" without clearly stating what P(k) says — always write out the assumption explicitly.
- Circular arrangements: forgetting to divide by n (i.e., using (n−1)! not n!).
Board Exam Tips
- Induction proofs are standard 5-mark questions — always follow all four steps explicitly.
- Combination problems about committees or groups (with conditions) appear as 3-mark questions.
- The sum formula (nC0 + nC1 + … + nCn = 2ⁿ) is useful in binomial theorem problems too.
- JEE: Circular permutation and combinations with restrictions are tested frequently.