Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Combinatorics and Mathematical Induction — Exercise Solutions
1. How to Use This Page
This page covers Chapter 4: Combinatorics and Mathematical Induction of the TN State Board Class 11 Maths textbook. Work through each section in order:
- Key Concepts (Quick Recap) — review formulas before attempting problems
- MCQs, Fill in the Blanks, True or False — test recall and understanding
- Match the Following — connect concepts to their formulas or values
- Answer Briefly — practise 2-mark reasoning questions
- Answer in Detail — master 5-mark problems with full worked solutions
- Common Mistakes to Avoid — spot and fix frequent errors
- Exam Checklist — ensure you have covered every exam topic
Every numerical solution is worked step by step. Try the problem on your own first, then compare your answer.
2. Key Concepts (Quick Recap)
Fundamental Principle of Counting
| Principle | Statement | Formula |
|---|---|---|
| Multiplication (AND) | If task A has m outcomes and task B has n outcomes, and both must be done | Total = m × n |
| Addition (OR) | If task A has m outcomes and task B has n outcomes, and either one suffices (mutually exclusive) | Total = m + n |
Factorial Notation
n! = n × (n − 1) × (n − 2) × … × 2 × 1
| Property | Value / Formula |
|---|---|
| 0! | 1 |
| 1! | 1 |
| n! | n × (n − 1)! |
| n! / (n − 1)! | n |
Permutations (nPr)
nPr = n! / (n − r)!
- Arrangements where order matters
- Number of permutations of n objects taken r at a time
| Special Case | Formula |
|---|---|
| nPn | n! |
| nP0 | 1 |
| Circular permutation of n distinct objects | (n − 1)! |
| Permutation with repetition of n objects where one repeats p times, another q times | n! / (p! × q!) |
| Arranging n objects with k alike of one type | n! / k! |
Combinations (nCr)
nCr = n! / (r! × (n − r)!)
- Selections where order does NOT matter
| Property | Formula |
|---|---|
| nC0 | 1 |
| nCn | 1 |
| nCr = nC(n−r) | Symmetry property |
| nCr + nC(r−1) = (n+1)Cr | Pascal's identity |
| nCr = nPr / r! | Relation with permutation |
Mathematical Induction
| Step | What to Do |
|---|---|
| Step 1: Base Case | Verify P(1) is true (or P(0) if starting from 0) |
| Step 2: Induction Hypothesis | Assume P(k) is true for some k ≥ 1 |
| Step 3: Inductive Step | Prove P(k + 1) is true using P(k) |
Key insight: Induction proves a statement is true for ALL natural numbers by proving a chain — if the first link holds and each link connects to the next, the entire chain holds.
3. I. Choose the Best Answer (MCQs)
Q1. The value of ⁿP₃ when n = 8 is:
⁸P₃ = 8! / (8 − 3)!
= 8! / 5!
= (8 × 7 × 6 × 5!) / 5!
= 8 × 7 × 6
= 336
Q2. If ⁿC₅ = ⁿC₇, then n is:
Using the property ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎:
ⁿC₅ = ⁿC₇
⟹ n − 5 = 7 or 5 = 7
⟹ n = 12
Q3. The number of ways 5 people can sit around a circular table is:
Circular permutation = (n − 1)! = (5 − 1)! = 4! = 24
Q4. If (n + 1)! = 720, then n equals:
(n + 1)! = 720
Since 6! = 720, we get n + 1 = 6
⟹ n = 5
Q5. The number of diagonals of a polygon with 10 sides is:
Number of diagonals = ⁿC₂ − n = ¹⁰C₂ − 10
= (10 × 9) / 2 − 10
= 45 − 10
= 35
Q6. The value of ¹²C₄ is:
¹²C₄ = 12! / (4! × 8!)
= (12 × 11 × 10 × 9) / (4 × 3 × 2 × 1)
= 11880 / 24
= 495
Q7. How many 3-digit numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?
⁵P₃ = 5! / (5 − 3)! = 5! / 2! = 120 / 2 = 60
4. II. Fill in the Blanks
1. ⁷P₀ = 1 (since nP0 = 1 for all n ≥ 0)
2. If ⁿC₂ = 45, then n = 10
ⁿC₂ = n! / (2! × (n − 2)!) = n(n − 1) / 2 = 45
⟹ n(n − 1) = 90
⟹ n = 10
3. The number of ways to choose 3 books from 8 is 56 (⁸C₃ = 56)
⁸C₃ = (8 × 7 × 6) / (3 × 2 × 1) = 336 / 6 = 56
4. ⁹C₃ + ⁹C₂ = ¹⁰C₃ (by Pascal's identity: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ)
5. 0! = 1
6. If ⁿP₂ = 90, then n = 10
ⁿP₂ = n! / (n − 2)! = n(n − 1) = 90
⟹ n² − n − 90 = 0
⟹ (n − 10)(n + 9) = 0
⟹ n = 10 (since n must be positive)
7. The number of permutations of the letters of the word "BANANA" is 60
BANANA has 6 letters: A appears 3 times, N appears 2 times
Number of arrangements = 6! / (3! × 2!) = 720 / 12 = 60
8. The number of 3-digit numbers that can be formed using digits 1, 2, 3, 4, 5 without repetition is 60 (⁵P₃ = 60)
9. ⁿCₓ = ⁿCᵧ implies x = y or x + y = n
10. The number of ways of selecting 4 players from 6 is 15 (⁶C₄ = 6!/(4! × 2!) = 15)
11. In mathematical induction, the step where we assume P(k) is true is called the induction hypothesis
12. The number of straight lines formed by joining 8 points, no three of which are collinear, is 28 (⁸C₂ = 28)
5. III. True or False (with corrections for false statements)
1. True or False: ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎
2. True or False: nPn = n!
3. True or False: In mathematical induction, verifying P(1) is optional if P(k) → P(k+1) is proved.
4. True or False: ⁿPᵣ = ⁿCᵣ × r!
5. True or False: The number of ways to arrange n distinct objects in a circle is n!.
6. True or False: ⁿC₁ + ⁿC₂ + … + ⁿCₙ = 2ⁿ − 1
7. True or False: Mathematical induction can be used to prove statements about natural numbers only.
6. IV. Match the Following
| Column A | Column B | Answer |
|---|---|---|
| (a) ⁵P₅ | (1) 1 | (a) → (3) 120 |
| (b) ⁶C₀ | (2) 210 | (b) → (1) 1 |
| (c) ¹⁰C₄ | (3) 120 | (c) → (2) 210 |
| (d) (n − 1)! in a circle | (4) nCr × r! | (d) → (5) Circular permutation |
| (e) nPr | (5) Circular permutation | (e) → (4) nCr × r! |
| (f) ⁿC₍ₙ₋ᵣ₎ | (6) ⁿCr | (f) → (6) ⁿCr |
Detailed working:
- (a) ⁵P₅ = 5! = 120 → matches (3)
- (b) ⁶C₀ = 1 → matches (1)
- (c) ¹⁰C₄ = 10! / (4! × 6!) = (10 × 9 × 8 × 7) / 24 = 210 → matches (2)
- (d) Circular permutation = (n − 1)! → matches (5)
- (e) nPr = nCr × r! → matches (4)
- (f) ⁿC₍ₙ₋ᵣ₎ = ⁿCr by symmetry → matches (6)
7. V. Answer Briefly (2-mark questions)
Q1. Prove that ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ
LHS = ⁿCᵣ + ⁿCᵣ₋₁
= n! / (r!(n − r)!) + n! / ((r − 1)!(n − r + 1)!)
Taking common denominator r!(n − r + 1)!:
= n! × [(n − r + 1) + r] / (r!(n − r + 1)!)
= n! × (n + 1) / (r!(n + 1 − r)!)
= (n + 1)! / (r!(n + 1 − r)!)
= ⁿ⁺¹Cᵣ
= RHS
Q2. How many 5-digit even numbers can be formed using digits 1, 2, 3, 4, 5 without repetition?
For the number to be even, the last digit must be 2 or 4 (2 choices).
Remaining 4 positions can be filled by the remaining 4 digits in 4! ways.
Total = 2 × 4! = 2 × 24 = 48
Q3. Find the number of ways to distribute 3 different prizes among 10 students if no student gets more than one prize.
First prize: 10 choices
Second prize: 9 choices (different student)
Third prize: 8 choices
By multiplication principle: 10 × 9 × 8 = 720 ways
Alternatively: ¹⁰P₃ = 720
Q4. State the Principle of Mathematical Induction.
If P(n) is a statement involving the natural number n such that:
1. P(1) is true (base case), and 2. P(k + 1) is true whenever P(k) is true (inductive step),
then P(n) is true for all natural numbers n.
Q5. Find r if ⁸Pᵣ = 8! and ⁸Pᵣ₋₁ = 7!
Given ⁸Pᵣ = 8!
⟹ 8! / (8 − r)! = 8!
⟹ (8 − r)! = 1
⟹ 8 − r = 0 (since 0! = 1)
⟹ r = 8
Alternatively, given both conditions:
⁸Pᵣ₋₁ = 7! means r − 1 = 1, so r = 2... but that gives ⁸P₂ = 56 ≠ 8!
So r = 8 is the answer: ⁸P₈ = 8! ✓
8. VI. Answer in Detail (5-mark questions)
Q1. Prove by mathematical induction that 1 + 2 + 3 + … + n = n(n + 1)/2 for all n ∈ ℕ.
Let P(n): 1 + 2 + 3 + … + n = n(n + 1)/2
Step 1 — Base Case (n = 1):
LHS = 1
RHS = 1 × (1 + 1) / 2 = 2/2 = 1
LHS = RHS ✓
Step 2 — Induction Hypothesis:
Assume P(k) is true for some k ≥ 1:
1 + 2 + 3 + … + k = k(k + 1)/2 … (i)
Step 3 — Inductive Step (Prove P(k + 1)):
We need to show: 1 + 2 + … + k + (k + 1) = (k + 1)(k + 2)/2
LHS = [1 + 2 + … + k] + (k + 1)
= k(k + 1)/2 + (k + 1) [using hypothesis (i)]
= [k(k + 1) + 2(k + 1)] / 2
= (k + 1)(k + 2) / 2
= RHS ✓
By the Principle of Mathematical Induction, P(n) is true for all n ∈ ℕ.
Q2. Prove by mathematical induction that n³ − n is divisible by 6 for all n ∈ ℕ.
Let P(n): n³ − n is divisible by 6.
Step 1 — Base Case (n = 1):
1³ − 1 = 0 = 6 × 0 ✓ (0 is divisible by 6)
Step 2 — Induction Hypothesis:
Assume P(k) is true: k³ − k = 6m for some integer m … (i)
Step 3 — Inductive Step:
Prove: (k + 1)³ − (k + 1) is divisible by 6.
(k + 1)³ − (k + 1)
= k³ + 3k² + 3k + 1 − k − 1
= k³ − k + 3k² + 3k
= (k³ − k) + 3k(k + 1)
From hypothesis (i): k³ − k = 6m (divisible by 6)
Now, k(k + 1) is the product of two consecutive integers,
which is always even, so 3k(k + 1) is divisible by 6.
Therefore: (k³ − k) + 3k(k + 1) = 6m + 6p (for some integer p)
= 6(m + p), which is divisible by 6 ✓
By the Principle of Mathematical Induction, n³ − n is divisible by 6 for all n ∈ ℕ.
Q3. A committee of 5 is to be formed from 7 men and 6 women. In how many ways can this be done if (i) exactly 3 women are on the committee, (ii) at least 3 women are on the committee?
Total people = 7 men + 6 women = 13
(i) Exactly 3 women (and therefore 2 men):
= ⁶C₃ × ⁷C₂
= (6 × 5 × 4) / (3 × 2 × 1) × (7 × 6) / (2 × 1)
= 20 × 21
= 420 ways
(ii) At least 3 women (3 or 4 or 5 women):
Case 1: Exactly 3 women, 2 men
= ⁶C₃ × ⁷C₂ = 20 × 21 = 420
Case 2: Exactly 4 women, 1 man
= ⁶C₄ × ⁷C₁ = 15 × 7 = 105
Case 3: Exactly 5 women, 0 men
= ⁶C₅ × ⁷C₀ = 6 × 1 = 6
Total = 420 + 105 + 6 = 531 ways
Q4. Prove by mathematical induction that 1² + 2² + 3² + … + n² = n(n + 1)(2n + 1)/6
Let P(n): 1² + 2² + 3² + … + n² = n(n + 1)(2n + 1)/6
Step 1 — Base Case (n = 1):
LHS = 1² = 1
RHS = 1 × 2 × 3 / 6 = 6/6 = 1
LHS = RHS ✓
Step 2 — Induction Hypothesis:
Assume: 1² + 2² + … + k² = k(k + 1)(2k + 1)/6 … (i)
Step 3 — Inductive Step:
Prove: 1² + 2² + … + k² + (k + 1)² = (k + 1)(k + 2)(2k + 3)/6
LHS = k(k + 1)(2k + 1)/6 + (k + 1)² [using (i)]
= (k + 1) × [k(2k + 1)/6 + (k + 1)]
= (k + 1) × [k(2k + 1) + 6(k + 1)] / 6
= (k + 1) × [2k² + k + 6k + 6] / 6
= (k + 1) × [2k² + 7k + 6] / 6
= (k + 1) × (k + 2)(2k + 3) / 6
Factorising 2k² + 7k + 6 = (k + 2)(2k + 3):
= (k + 1)(k + 2)(2k + 3) / 6 ✓
By the Principle of Mathematical Induction, the formula holds for all n ∈ ℕ.
Q5. How many words (with or without meaning) can be formed using all the letters of the word "MISSISSIPPI"? In how many of these do the four I's not come together?
The word MISSISSIPPI has 11 letters:
M = 1, I = 4, S = 4, P = 2
Total letters = 11
(i) Total arrangements:
= 11! / (4! × 4! × 2!)
= 39916800 / (24 × 24 × 2)
= 39916800 / 1152
= 34650
(ii) When all four I's are together:
Treat IIII as a single letter. Then we have:
M, S, S, S, S, P, P, (IIII) — 8 objects
where S = 4 alike, P = 2 alike
Arrangements = 8! / (4! × 2!)
= 40320 / (24 × 2)
= 40320 / 48
= 840
(iii) I's NOT together = Total − I's together
= 34650 − 840
= 33810
9. Common Mistakes to Avoid
| # | Mistake | Correction |
|---|---|---|
| 1 | Confusing permutation with combination — using nCr when order matters | Use nPr when arranging (order matters), nCr when selecting (order does not matter) |
| 2 | Writing nCr = n! / (n − r)! instead of n! / (r!(n − r)!) | nCr has r! in the denominator; nPr does not. The formula for nCr includes dividing by r! |
| 3 | Using n! for circular permutations instead of (n − 1)! | Circular arrangements use (n − 1)! because rotating all objects preserves the relative order |
| 4 | Forgetting to subtract from total in "at least" problems | "At least r" means r, r+1, r+2, … — sum all cases or use Total − (unwanted cases) |
| 5 | Assuming 0! = 0 instead of 0! = 1 | 0! = 1 by convention; this is essential for formulas like nC0 = n! / (0! × n!) = 1 |
| 6 | In induction, assuming P(k) is true for ALL k instead of a SPECIFIC k | The hypothesis assumes P(k) is true for ONE specific arbitrary k, then proves P(k+1) follows |
| 7 | Skipping the base case in induction | Without verifying P(1) (or P(0)), the inductive step proves nothing — the chain has no starting point |
| 8 | Using nPn = n for permutations of all objects | nPn = n! (not n). Permuting all n distinct objects gives n! arrangements |
| 9 | Confusing "and" vs "or" in counting problems | "And" → multiply (both must happen). "Or" → add (either can happen, if mutually exclusive) |
| 10 | Not accounting for identical objects in permutation | When objects repeat, divide by the factorial of each repetition count: n! / (p! × q! × …) |
| 11 | Forgetting that product of consecutive integers is divisible by 2, and product of three consecutive integers by 6 | n(n+1) is always even; n(n+1)(n+2) is always divisible by 6 — useful in induction proofs |
| 12 | Not checking if r ≤ n before applying nPr or nCr | nPr and nCr are defined only when 0 ≤ r ≤ n. If r > n, the value is 0 |
10. Exam Checklist
Before entering the exam hall, make sure you can do the following without looking at notes:
- [ ] State the Fundamental Principle of Counting (multiplication and addition) and identify when to use each
- [ ] Compute factorials quickly for n ≤ 10 (know that 0! = 1, 5! = 120, 10! = 3628800)
- [ ] Distinguish permutation vs combination — if the problem asks "arrange", "order", "rank", "line up" → permutation; if it asks "choose", "select", "pick", "team" → combination
- [ ] Apply nPr and nCr formulas correctly, including the relation nPr = nCr × r!
- [ ] Handle circular permutations — remember (n − 1)!, and for necklaces/bracelets divide further by 2
- [ ] Count arrangements with restrictions — vowels together (treat as block), vowels never together (total − together), specific positions (fix those first)
- [ ] Apply the "at least / at most" technique — sum individual cases or subtract from total
- [ ] Execute a complete mathematical induction proof — base case, hypothesis, inductive step clearly labelled
- [ ] Prove summation formulas by induction — sum of first n naturals, sum of squares, sum of cubes
- [ ] Prove divisibility by induction — e.g., n³ − n divisible by 6, 2ⁿ > n for n ≥ 1
- [ ] Use symmetry property nCr = nC(n−r) to simplify calculations
- [ ] Use Pascal's identity nCr + nC(r−1) = (n+1)Cr when combining selections
- [ ] Check dimensional consistency in your answer — if choosing 3 from 10, the answer must be ≤ 10C3 = 120
- [ ] Know standard results: number of diagonals of an n-sided polygon = ⁿC₂ − n; number of triangles from n non-collinear points = ⁿC₃
Frequently Asked Questions
Use permutation (nPr) when order matters; use combination (nCr) when order does not matter.