Quantitative Aptitude
Permutation and Combination; Probability
C-CAT
Permutation and Combination
19.1 Fundamental Counting Principle
If task 1 can be done in m ways and task 2 in n ways:
- If BOTH needed (AND): m × n ways (Multiplication)
- If EITHER needed (OR): m + n ways (Addition)
19.2 Factorial
n! = n × (n-1) × (n-2) × ... × 2 × 1
Special cases: 0! = 1, 1! = 1
19.3 Permutation (Order Matters)
ⁿPᵣ = n! / (n-r)!
Reads: "n choose r with order"
Example: ⁵P₃ = 5!/(5-3)! = 5!/2! = 60
When items are repeated:
- Permutations of n items where a items are same, b items are same:
= n! / (a! × b!)
Example: MATHEMATICS (11 letters, M×2, A×2, T×2):
- = 11! / (2! × 2! × 2!) = 4,989,600
19.4 Combination (Order Doesn't Matter)
ⁿCᵣ = n! / [r! × (n-r)!]
Key Properties:
- ⁿCᵣ = ⁿC(n-r)
- ⁿC₀ = ⁿCₙ = 1
- ⁿC₁ = n
- ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ (Pascal's Identity)
19.5 Permutation vs Combination
| Situation | Use |
|---|---|
| Arrangement (order matters) | Permutation |
| Selection (order doesn't matter) | Combination |
| Passwords, rankings, words | Permutation |
| Committees, teams, groups | Combination |
| Sitting arrangements (line) | Permutation |
| Sitting arrangements (circle) | (n-1)! |
19.6 Special Cases
Circular Arrangement
- n objects in a circle = (n-1)! ways
- n objects on a necklace (both sides identical) = (n-1)! / 2
Choosing with Restrictions
At least / at most problems: Use complementary counting
- At least 1 = Total - (choosing
- At most 2 = Choose 0 + Choose 1 + Choose 2
Example: From 6 men 4 women, committee of 5 with at least 2 women:
- = (4C2)(6C3) + (4C3)(6C2) + (4C4)(6C1) + (4C5)(6C0) Wait: at least 2 women:
- 2W + 3M = 4C2 × 6C3 = 6 × 20 = 120
- 3W + 2M = 4C3 × 6C2 = 4 × 15 = 60
- 4W + 1M = 4C4 × 6C1 = 1 × 6 = 6
- Total = 186
Words from a Word (vowels together)
- Treat vowels as a single unit
- Arrange (consonants + 1 unit) + arrange vowels within unit
Example: OPTICAL (7 letters, vowels O, I, A):
- Vowels together: treat OIA as 1 block → 5! arrangements
- Vowels within block: 3! = 6
- Total = 5! × 3! = 120 × 6 = 720
19.7 nPr and nCr Quick Lookups
| n | r | nPr | nCr |
|---|---|---|---|
| 5 | 2 | 20 | 10 |
| 5 | 3 | 60 | 10 |
| 6 | 2 | 30 | 15 |
| 6 | 3 | 120 | 20 |
| 10 | 2 | 90 | 45 |
| 10 | 3 | 720 | 120 |
| 50 | 2 | 2450 | 1225 |
Probability
20.1 Basic Definition
Probability = Favorable Outcomes / Total Outcomes
- P(E) ranges from 0 to 1
- P(sure event) = 1
- P(impossible event) = 0
- P(E) + P(E') = 1 → P(E') = 1 - P(E) (Complementary events)
20.2 Sample Spaces
Coin Toss
- 1 coin: S = {H, T}, n(S) = 2
- 2 coins: S = {HH, HT, TH, TT}, n(S) = 4
- 3 coins: n(S) = 8
- n coins: n(S) = 2ⁿ
Dice Roll
- 1 die: S = {1,2,3,4,5,6}, n(S) = 6
- 2 dice: n(S) = 36
Card Deck (52 cards)
- 4 suits: Spades ♠, Hearts ♥, Diamonds ♦, Clubs ♣
- Each suit: 13 cards (A, 2-10, J, Q, K)
- Red cards: Hearts + Diamonds = 26
- Black cards: Spades + Clubs = 26
- Face cards: J, Q, K in each suit = 12 total
- Aces: 4 total
20.3 Key Probability Formulas
| Event | Formula |
|---|---|
| P(A or B) — Union | P(A) + P(B) - P(A∩B) |
| P(A and B) — Intersection | P(A) × P(B) [if independent] |
| P(A | B) — Conditional |
| Mutually exclusive events | P(A∩B) = 0, P(A∪B) = P(A) + P(B) |
| Independent events | P(A∩B) = P(A) × P(B) |
20.4 Common Probability Examples
Drawing from a bag of balls:
- P(red) = (number of red balls) / (total balls)
Drawing cards:
- P(face card) = 12/52 = 3/13
- P(king of red) = 2/52 = 1/26
- P(queen or king) = 8/52 = 2/13
Rolling dice:
- P(sum = 7 with 2 dice) = 6/36 = 1/6 (pairs: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1)
- P(at least one six) = 1 - P(no six) = 1 - (5/6)² = 11/36
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