1. Chapter at a Glance
How many 4-digit PINs are possible? In how many ways can 5 students sit in a row? How many cricket teams of 11 can be chosen from 15 players? Listing every possibility is impossible for big numbers, so we need clever counting. This chapter gives the fundamental principle of counting, permutations (arrangements, where order matters) and combinations (selections, where order does not matter). Indian mathematicians studied these ideas early — Bhāskarācārya's Līlāvatī (12th century) discusses counting arrangements — and they are the foundation of probability and the binomial theorem.
- Fundamental principle of counting (multiplication principle).
- Factorial notation n!.
- Permutations ⁿPᵣ of distinct objects; with repetition; with some objects alike.
- Combinations ⁿCᵣ and their properties.
- Deciding whether a problem needs permutations or combinations.
- Mixed problems: words, committees, digits, restrictions.
2. Fundamental Principle of Counting
If one event can occur in m ways and, after it, another independent event can occur in n ways, then both together can occur in m × n ways. This extends to any number of events.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if digits can be repeated?
Solution: Units place: 3 choices (2, 4, 6); tens: 6; hundreds: 6. Total = 6 × 6 × 3 = 108.
3. Factorial
n! = 1 × 2 × 3 × … × n, with 0! = 1. Also n! = n × (n − 1)!.
| n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|---|
| n! | 1 | 1 | 2 | 6 | 24 | 120 | 720 | 5040 |
4. Permutations
The number of arrangements of n distinct objects taken r at a time (0 ≤ r ≤ n), without repetition:
All n objects: ⁿPₙ = n!.
| Situation | Count |
|---|---|
| n distinct, r at a time, repetition allowed | nʳ |
| n objects with p alike of one kind, q of another, … | n! / (p! q! …) |
How many different words (with or without meaning) can be formed from all the letters of ALLAHABAD?
Solution: 9 letters: A × 4, L × 2, H, B, D. Number = 9!/(4! 2!) = 362880/48 = 7560.
In how many arrangements of the letters of MONDAY do the vowels occupy only the first and last places?
Solution: Vowels O, A in the two ends: 2! = 2 ways. Remaining 4 letters in the middle: 4! = 24. Total = 48.
In how many ways can 5 boys and 3 girls sit in a row so that all girls sit together?
Solution: Treat the 3 girls as one block: 6 units → 6!. Girls within the block: 3!. Total = 720 × 6 = 4320.
5. Combinations
The number of ways of choosing r objects from n distinct objects (order does not matter):
Each selection of r objects can be arranged in r! ways, which is why ⁿPᵣ = r! × ⁿCᵣ.
- ⁿC₀ = ⁿCₙ = 1; ⁿC₁ = n.
- ⁿCᵣ = ⁿCₙ₋ᵣ (choosing r is the same as rejecting n − r).
- ⁿCₐ = ⁿC_b ⇒ a = b or a + b = n.
- ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's rule).
A committee of 3 men and 2 women is to be chosen from 5 men and 4 women. In how many ways?
Solution: ⁵C₃ × ⁴C₂ = 10 × 6 = 60.
A committee of 7 is chosen from 9 boys and 4 girls with at least 3 girls. In how many ways?
Solution: 3 girls: ⁴C₃ × ⁹C₄ = 4 × 126 = 504. 4 girls: ⁴C₄ × ⁹C₃ = 84. Total = 588.
How many triangles can be formed from 12 points in a plane, no three collinear? How many lines?
Solution: Triangles ¹²C₃ = 220; lines ¹²C₂ = 66.
6. Permutation or Combination?
| Question words | Use | Example |
|---|---|---|
| Arrange, order, line up, rank, digits, words, seating | Permutation | President and secretary from 10 people: ¹⁰P₂ = 90 |
| Choose, select, committee, team, group, handshakes | Combination | 2 representatives from 10: ¹⁰C₂ = 45 |
| Select then arrange | Both: ⁿCᵣ × r! | Words using 2 vowels and 3 consonants |
Chart — ¹⁰Cᵣ for r = 0 to 5 (symmetric for r = 5 to 10):
7. Solved Problems (Board Pattern)
Find n if ⁿ⁻¹P₃ : ⁿP₄ = 1 : 9.
Solution: (n − 1)(n − 2)(n − 3) / [n(n − 1)(n − 2)(n − 3)] = 1/n = 1/9 → n = 9.
If ⁿC₈ = ⁿC₂, find ⁿC₂.
Solution: 8 + 2 = n → n = 10. ¹⁰C₂ = 45.
How many words, each of 3 vowels and 2 consonants, can be formed from the letters of INVOLUTE?
Solution: Vowels I, O, U, E (4); consonants N, V, L, T (4). Choose ⁴C₃ × ⁴C₂ = 24; arrange each 5-letter selection in 5! = 120 ways. Total = 2880.
8. Master Formula Table
| Count | Formula |
|---|---|
| Multiplication principle | m × n × … |
| Arrangements, no repetition | ⁿPᵣ = n!/(n − r)! |
| Arrangements, repetition allowed | nʳ |
| Arrangements with identical objects | n!/(p! q! r! …) |
| Selections | ⁿCᵣ = n!/(r!(n − r)!) |
| Link | ⁿPᵣ = r! ⁿCᵣ |
| Symmetry | ⁿCᵣ = ⁿCₙ₋ᵣ |
| Pascal's rule | ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ |
| Diagonals of n-gon | ⁿC₂ − n |
9. Master Summary Table
| Situation | Approach | Typical question |
|---|---|---|
| Independent choices | Multiply | Number plates, PINs, outfits |
| Order matters | Permutation | Words, numbers, seating, ranks |
| Order doesn't matter | Combination | Teams, committees, triangles |
| Items together | Treat as one block | Girls together, vowels together |
| Items never together | Total − together, or gaps method | No two vowels adjacent |
| "At least / at most" | Add cases or use complement | Committee with ≥ 3 girls |
10. Mind Map
- Multiply choices
- n!
- ⁿPᵣ
- Alike objects
- ⁿCᵣ
- Symmetry, Pascal
- Restrictions first
- Cases, complement
11. Tips and Tricks
- Fill restricted positions first (e.g. the units digit for even numbers, the first digit ≠ 0).
- For "never together", place the other items first and put the restricted ones in the gaps.
- Use ⁿCᵣ = ⁿCₙ₋ᵣ to compute with the smaller r: ¹⁰⁰C₉₈ = ¹⁰⁰C₂ = 4950.
- Using permutations for a committee (order does not matter).
- Forgetting to divide by p! for repeated letters.
- Allowing 0 as the first digit of a number.
12. Important One-liners
- 0! = 1.
- ⁿPᵣ counts arrangements; ⁿCᵣ counts selections.
- ⁿPᵣ = r! × ⁿCᵣ.
- ⁿCᵣ = ⁿCₙ₋ᵣ.
- The number of handshakes among n people is ⁿC₂.
- Arrangements of n objects with p alike = n!/p!.
13. Exam-pattern Questions with Detailed Answers
The value of 8!/(6! × 2!) is:
- 28
- 56
- 14
- 336
Show answer
(8 × 7)/2 = 28.
Answer: (A)
How many 4-letter codes can be formed using the first 10 letters of the English alphabet if no letter is repeated?
Show answer
¹⁰P₄ = 10 × 9 × 8 × 7 = 5040.
Answer: 5040
In how many ways can a team of 3 be chosen from 8 players?
Show answer
⁸C₃ = 56.
Answer: 56
How many 5-digit telephone numbers can be formed using 0–9 if each number starts with 67 and no digit appears more than once?
Show answer
Remaining 3 places from 8 digits: ⁸P₃ = 336.
Answer: 336
How many arrangements of the letters of MISSISSIPPI are there?
Show answer
11 letters: S × 4, I × 4, P × 2, M. 11!/(4! 4! 2!) = 34650.
Answer: 34650
How many diagonals does a polygon with 12 sides have?
Show answer
¹²C₂ − 12 = 66 − 12 = 54.
Answer: 54
Determine n if ²ⁿC₃ : ⁿC₃ = 12 : 1.
Show answer
[2n(2n − 1)(2n − 2)]/[n(n − 1)(n − 2)] = 12 → 4(2n − 1)/(n − 2) = 12 → 8n − 4 = 12n − 24 → n = 5.
Answer: n = 5
In how many ways can 5 cards be selected from a deck of 52 so that there is exactly one ace?
Show answer
⁴C₁ × ⁴⁸C₄ = 4 × 194580 = 778320.
Answer: 778320
In how many ways can the letters of PERMUTATIONS be arranged if (i) words start with P and end with S, (ii) vowels are all together?
Show answer
12 letters with T × 2. (i) Middle 10 letters (T twice): 10!/2! = 1814400. (ii) Vowels E, U, A, I, O as one block: 8 units (T twice) → 8!/2! = 20160; vowels inside 5! = 120 → 2419200.
Answer: 1814400 and 2419200
In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if discs of the same colour are identical?
Show answer
9!/(4! 3! 2!) = 1260.
Answer: 1260
In how many ways can 6 boys and 4 girls be seated in a row so that no two girls sit together?
Show answer
Seat boys: 6! = 720. They create 7 gaps; choose and arrange 4 girls: ⁷P₄ = 840. Total = 604800.
Answer: 604800
A student must answer 10 of 13 questions, choosing at least 4 from the first 5. In how many ways?
Show answer
4 from first 5 and 6 from remaining 8: ⁵C₄ × ⁸C₆ = 5 × 28 = 140. All 5 and 5 from 8: 1 × 56 = 56. Total 196.
Answer: 196
14. Quick Revision Summary
- Multiply the numbers of choices for independent steps.
- ⁿPᵣ = n!/(n − r)! for arrangements; nʳ with repetition; n!/(p!q!…) with identical objects.
- ⁿCᵣ = n!/(r!(n − r)!) for selections; ⁿCᵣ = ⁿCₙ₋ᵣ; Pascal's rule.
- Order matters → P; order doesn't → C; select-then-arrange → C × r!.
- Handle restrictions first; use blocks for "together", gaps for "not together".